Class 10 Maths

Quadratic Equations

Question 1. Which of the following are quadratic equations?

1.  X²–8x+12=0

Solution:

Given Solution is x²-8X+12=0

⇒  x²-6x-2x+12=0

⇒  x(x-6)-2(x-6) = 0

⇒ (x-2)(x-6)=0

⇒ x-2=0 or 2-6=0

⇒ x=2 Or x=6

Hence, x=2 and x=6 are the solutions.

2. 5x²-7x=3x²-7x+3

Solution:

Given Solution is 5x²-7x=3x²-7x+3

5x²-7x-3x²+7x-3=0

2x²-3=0

2x²=3

x² = \(\frac{3}{2}\)

⇒ \(x=\sqrt{\frac{3}{2}}\)

Hence , \(x=\sqrt{\frac{3}{2}}\) are the solutions.

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3. \(\frac{1}{4} x^2+\frac{7}{6} x-2=0\)

Solution:

Given equation is \(\frac{1}{4} x^2+\frac{7}{6} x-2=0\)

⇒ \(\frac{3 x^2+14 x-24}{12}=0\)

3x²+ 14x-24= 0 (1)

Equation in the form od ax² + bx + c =0

a = 3, b = 14, c = -24

⇒ \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(\frac{-14 \pm \sqrt{(14)^2-4(3)(-24)}}{2(3)}\)

⇒ \(\frac{-14 \pm \sqrt{196+288}}{6}\)

⇒ \(\frac{-14 \pm \sqrt{484}}{6}\)

⇒ \(\frac{-14 \pm 22}{6}\)

⇒ \(\frac{-14+22}{6} \text { or } \frac{-14-22}{6}\)

⇒ \(\frac{8}{6} \text { or } \frac{-36}{6}\)

⇒ \(\frac{4}{3} \text { or }-6\)

Hence \(x=\frac{4}{3} \text { and } x=-6\) are the solutions.

Question 2. Which of the following are roots of 4x²-9x-100=0?

1. -4
2. \(\frac{3}{4}\)
3. \(\frac{25}{4}\)

Solution:

The given equation is 4x²-9x-100=0

On substituting x=-4 in the given equation

L.H.S= 4(-4)²-9(-4) – 100 = 0

64+34-100= 0 ⇒ = 0 = R.H.S

∴ x=-4 is a Solution of 4×2-9x-100.

On Substituting x=2/14 in the given equation

L.H.S= \(4\left(\frac{3}{4}\right)^2-9\left(\frac{3}{4}\right)-100=0\)

⇒ \(4\left(\frac{9}{16}\right)-\frac{27}{4}-100=0\)

⇒ \(\frac{36-108-1000}{16}=0\)

= 36-208=0

= 178 not equal to R.H.S

⇒ \(x=\frac{3}{4}\) is not a solutions of 4×2 9x – 100=0

On substituting \(x=\frac{25}{4}\)

⇒ \(\text { L.H.S }=4\left(\frac{25}{4}\right)^2-9\left(\frac{25}{4}\right)-100=0\)

⇒ \(4\left(\frac{625}{16}\right)-\frac{225}{4}-100=0\)

⇒ \(\frac{2500-900-1600}{16}=0\)

2500-2500=0 = R.H.S

⇒ \(x=\frac{25}{4}\) is a solution of 4×2 is a solution.

Question 3. If one root of the quadratic equation 6x²-x-k=0 is 2, find the k value.

Solution:

Since, \(x=\frac{2}{3}\) is a solution of 6x²-x-k=0

⇒ \(6\left(\frac{2}{3}\right)^2-\frac{2}{3}-k=0\)

⇒ \(6\left(\frac{4}{9}\right)-\frac{2}{3}-k=0\)

⇒ \(\frac{24-6-9 k}{9}=0\)

18-9k = 0

18= 9k

⇒ \(k=\frac{18}{9}\)

k=2

Hence k=2 of the solution.

Question 4. 3x²-243=0

Solution:

Given equation is 3x²-243=0

3x²=243

⇒ \(x^2=\frac{243}{3}\)

x⇒ = 81

⇒ \(x=\sqrt{81}\)

⇒ \(x= \pm 9\)

x = 9 or x = -9

Hence x = 9 and x = -9 are the solutions.

Question 5. 5x²+4x=0

Solution:

Given equation is 5x²+4x=0

x(5x+4)= 0

x = 0 or 5x+4 – 0

5X=-4

x= \(\frac{-4}{5}\)

Hence x=0 and x = \(\frac{-4}{5}\) are the Solutions.

Question 6. x² +12x+35=0

Solution:

The given equation is x² +12x+35=0

x² +5x+7x+35=0

x(x+5)+7(x+5)=

(1+7)(x+5)=0

x+7=0 or x+5=0

x=-7 or x=-5

Hence 2=-7 and x=-5 are the solutions.

Question 7. 2x²=5x+3=0

Solution:

Given equation is 2x²-5x+3=0

2x² =5x+3=0

2x² =3x-2x+3=0

x(2x-3)-1(2x-3)=0

(x-1)(2x-3)=0

x-1=0 or 2x-3=0

x=1 or 2x-3=0

2x=3

⇒ x = \(=\frac{3}{2}\)

Hence x=1 and x = \(=\frac{3}{2}\) are the Solutions.

Question 8. 6x²-x-2=0

Solution:

Given equation is 6x²-x-2=0

⇒ 6x²-4x+3x-2=0

⇒ 2x(3x-2)+1(3x-2)=0

⇒ (x+1)(3x-2)=0

⇒ 2x+1=0 Or 3x-2=0

⇒ 2x=-1 or 3x=2

⇒ \(x=-\frac{1}{2}\) Or \(x=\frac{2}{3}\)

Hence \(x=-\frac{1}{2}\) and \(x=\frac{2}{3}\) are the Solutions.

Question 9. 8x²-2x-21=0

Solution:

Given equation are 8x²-2x-21=0

8x²+6x-28x-21=0

2x(4x+3)-7(4x+3)=0

(2x-7)(4x+3)= 0

2x-7=0 4x+3=0

2x=7 Οr 4x=-3

⇒ \(x=\frac{7}{2}\) or \(x=\frac{-3}{4}\)

Hence \(x=\frac{7}{2}\) and \(x=\frac{-3}{4}\)

Question 10. 6x+40=31x

Solution:

Given equation are 6×2-31x+40=0

⇒ 62x= 18x-16x+40=0

⇒ 3x(2x-5)-8(2x-5)=6

⇒ (3x-8)(2x-5)=6

⇒ 3x-8=0 or 2x-5=0

⇒ 3x=8 Οr 2x = 5

⇒ \(x=\frac{8}{3}\) Or \(x=\frac{5}{2}\)

Hence \(x=\frac{8}{3}\) and \(x=\frac{5}{2}\) are the Solutions:

Question 11. \(\sqrt{3} x^2-11 x+8 \sqrt{3} x=0\)

Solution:

Given Equation is √3x²=11x+8√3=0

Equation in the form ax+ bx+c

a=√3, b=-11, C=8√3

⇒ \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(\frac{11 \pm \sqrt{(-11)^2-4(\sqrt{3})(8 \sqrt{3})}}{2 \sqrt{3}}\)

⇒ \(\frac{11 \pm \sqrt{121-96}}{2 \sqrt{3}}\)

⇒ \(\frac{11 \pm \sqrt{25}}{2 \sqrt{3}}\)

⇒ \(\frac{11 \pm 5}{2 \sqrt{3}}\)

⇒ \(\frac{11+5}{2 \sqrt{3}} \text { or } \frac{11-5}{2 \sqrt{3}}\)

⇒ \(\frac{16}{2 \sqrt{3}} \text { or } \frac{6}{2 \sqrt{3}}\)

⇒ \(\frac{8}{\sqrt{3}} \text { or } \frac{3}{\sqrt{3}}\)

⇒ \(\frac{8 \times 3}{\sqrt{3}} \text { or } \sqrt{3}\)

⇒ \(8 \sqrt{3}\)

x=8√3 and x =√3

Hence x=8√3 and x =√3 are the solutions.

Question 12.3x² – 256x+2=0

Solution:

Given equation is 3x²-2√6x+2=0

Equation in the form of an 7 bx + c = 0

a = 3, 6=-256, C= 2

⇒ \(\Rightarrow \frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(\frac{2 \sqrt{6} \pm \sqrt{(-2 \sqrt{6})^2-4(3)(2)}}{2(3)}\)

⇒ \(\frac{2 \sqrt{6} \pm \sqrt{26-26}}{6}\)

⇒ \(\frac{2 \sqrt{6}}{6}\)

⇒ \(\frac{2}{\sqrt{6}}\)

⇒ \(\frac{2}{\sqrt{2 \times 3}}\)

⇒ \(\frac{\sqrt{2}}{\sqrt{3}} \Rightarrow \sqrt{\frac{2}{3}}\)

Hence \(x=\sqrt{\frac{2}{3}}\) are the solution.

Question 13. x² + 5= \(\frac{9}{2} x\)

Solution:

Given equation is x² + 5 – \(\frac{9}{2} x\)

2x² +10-9x=0

⇒ 2x²=9x+10=0

⇒ 2x²=4x-5x+10=0

⇒ 2x(x-2)-5(x-2)=0

⇒ (2x-5)(x-2)=0

⇒ 2x-5=0 or x-2=0

⇒ 2x=5 or x=2

⇒ \(x=\frac{5}{2}\)

Hence \(x=\frac{5}{2}\) and x= 2 are the solution

Question 14. \(x=\frac{3 x+x}{}\)

Solution:

Given equation is x= \(x=\frac{3 x+x}{}\)

4x^2-3x+1

4x^2-37-1=0

4x²=4x-x-1=0

4x(x-1)+(x-1)=6

(4x+1)(x-1)=0

4×71=0 or x-1=0

4x=-1 or x=1

⇒  \(x=\frac{-1}{4}\)

Hence \(x=\frac{-1}{4}\) and x=1 are the solution.

Question 15. \(5 x-\frac{35}{x}=18, x \neq 0\)

Solution:

Given equation is \(5 x-\frac{35}{x}=18\)

5×2-18x-35=0

⇒ 5×2+7x-25x-35=0

⇒ x(5x+7)-5(5x+7)= 0

⇒ x-5=0 or 5x+7=0

⇒ x = 5 or 5x=-7

⇒ \(x=\frac{-7}{5}\)

Hence x = 5 and \(x=\frac{-7}{5}\) are the solution.

Question 16. \(\frac{2}{x^2}-\frac{5}{x}+2=0, x \neq 0\)

Solution:

Given equation is \(\frac{2}{x^2}-\frac{5}{x}+2=0\)

\(\frac{2-5 x+2 x^2}{x^2}=0\)

2x²-5x+2=0

2x²-4x-x+2=0

2x²-4x-x+2=0

2×(x-2)-(X-2) = 0

(2x-1)(x-2)=0

2x-1=0 or x=2=0

2x = 1 Or x = 2

⇒ \(x=\frac{1}{2}\)

Hence \(x=\frac{1}{2}\)or x=2 are the solution.

Question 17. a²x²+2ax+1= 0

Solution:

Given equation is a²x²+2ax+1= 0

a²x²+2ax+1= 0

a²x²+ax+ax+1= 0

ax(ax+1)+(ax+1)=0

(ax+1)(ax+1)=0

ax+1=0 or ax+1=0

ax=-1 Or ax = -1

⇒ \(x=\frac{-1}{a} \text { or } x=\frac{-1}{a}\)

Hence \(x=\frac{-1}{a} \text { or } x=\frac{-1}{a}\) are the solution.

Question 18. x² – (p+q)x+pq=0

Solution:

Given equation is x² – (p+q)x+pq=0

x² – qx – Px +Pq=0

x(x-2)-P(x-2)=0

(X-P) (x-2)=0

X-P=O or x-2=0

X=P or x=2

Hence x=P and x=q are the solutions.

Question 19. 12abx²-(9a²-8b²)x-6ab=0

Solution:

Given Equation is 12 abx² (9a²-8b²)x-6ab=0

12 abx = 9ax+8b-x=6ab=0

3ax (4bx-3a)+2b (4bx-3a)=0

(3ax+26) (46x-3a)=

3ax+2b=0 or 4bx-3a=0

3ax=-2b or 4bx=3a

Hence x = -2b and n=39 are the solutions.

⇒ \(x=\frac{-2 b}{3 a}\) Or \(x=\frac{3 a}{4 b}\)

Hence \(x=\frac{-2 b}{3 a}\) and \(x=\frac{3 a}{4 b}\) are the solution.

Question 20. 4x²-4ax+(a²-b²)=0

Solution:

Given Equation is 4x² – 4ax + (a² – b²) =0

4x² – 4ax + (a²-b²)=0

4x² = 49x + a² = b² = 0

⇒ \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(\frac{4 a \pm \sqrt{(4 a)^2-4(4)\left(a^2-b^2\right)}}{2(4)}\)

⇒ \(\frac{4 a \pm \sqrt{16 a^2-16 a^2+16 b^2}}{8}\)

⇒ \(\frac{4 a \pm \sqrt{(4 b)^x}}{8}\)

⇒ \(\frac{4 a+4 b}{8}\)

⇒ \(\frac{4 a+4 b}{8} \text { or } \frac{4 a-4 b}{8}\)

⇒ \(\frac{4(a+b)}{8} \text { or } \frac{4(a-b)}{8}\)

⇒ \(\frac{a+b}{2} \text { or } \frac{a-b}{2}\)

Hence \(x=\frac{a+b}{2} \text { and } x=\frac{a-b}{2}\) are the solution.

Find the roots of the following quadratic equations by the method of  Completing the Square.

Question 21. Find the roots of the following quadratic equations by the method of  Completing the Square x²=10x-24=0

Solution:

Given equation is x²-10x-24=0

on Comparing with ax²+ bx+ c=0, we get

a=1, b=-10, c= -24

Discriminant, D= b² – 4ac

⇒ D= (-10)2- 4(1)(-24)

⇒ D= 100+96

⇒ D = 196

Hence, the given equation has two real roots.

⇒ \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{+10 \pm \sqrt{196}}{2}\)

⇒ \(x=\frac{10 \pm 14}{2}\)

⇒ \(x=\frac{10+14}{2} \text { or } \frac{10-14}{2} \text {, }\)

⇒ \(x=\frac{24}{2} \text { or }-\frac{4}{2}\)

x= 12 Or – 2

x= 12,-2 are the roots of the equation.

Question 22. 2x²-7x-39=0

Solution:

The given equation is 2x²-7x-39=0

on Comparing that ax + bx + C=0, we get

= 2, 6=-7, C=-39

Discriminant D= b²-4ac

⇒ D= (-7)==4(2)(-39)

⇒ D= 49+312

⇒ D = 361

Hence, the given equation has two real roots

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{7 \pm \sqrt{361}}{2(2)}\)

⇒ \(x=\frac{7+19}{4} \text { or } \frac{7-19}{4}\)

⇒ \(x=\frac{13}{2} \text { or }-3\)

⇒ \(x=\frac{13}{2} \text { or }-3\) are roots of the equation.

Question 23. 5x² + 6x – 8 = 0

Solution:

Given equation is 5×2+6x-8 = 0

on Comparing that ax2+ bx+c=0, we get

a=5, b=6, C=-8

Discriminant D= b²– 4ac

⇒ D= (6)2-4(5)(-8)

⇒ D = 36 + 160

⇒ D= 196

Hence, the given equation has two real roots

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{-6 \pm \sqrt{196}}{2(5)}\)

⇒ \(x=\frac{-6 \pm 14}{10}\)

⇒ \(x=\frac{-6+14}{10} \text { or } \frac{-6-14}{10}\)

⇒ \(x=\frac{4}{5} \text { or }-2\)

⇒ \(x=\frac{4}{5} \text { or }-2\) are the real roots.

Question 24. \(\sqrt{3} x^2+11 x+6 \sqrt{3}=0\)

Solution:

Given Equation is \(\sqrt{3} x^2+11 x+6 \sqrt{3}=0\)

a=√3, b=11, C=6√3

Discriminant D =  b2 – 4ac

⇒ D= (11)² – 4(√3)(6√3)

⇒ D= 121-72

⇒ D = 49

Hence the given equation has two real roots

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{-11 \pm \sqrt{49}}{2 \sqrt{3}}\)

⇒ \(x=\frac{-11 \pm 7}{2 \sqrt{3}}\)

⇒ \(x=\frac{-11+7}{2 \sqrt{3}} \text { or } \frac{-11-7}{2 \sqrt{3}}\)

⇒ \(x=\frac{-4}{2 \sqrt{3}} \text { or } \frac{-18}{2 \sqrt{3}}\)

⇒ \(x=\frac{-2 \sqrt{3}}{3} \text { or }-3 \sqrt{3}\)
are the roots.

Question 25. 2x² -9x+7=0

Solution:

Given equation is 2x² = 9x + 7 = 0

2x² – 9x + 7 = 0

on Comparing that Qx²+6x+=0, we get

a=2, b = -9, c=7

Discriminant D = b² – 4aC

⇒ D = (-9)² – 4(2)(7)

⇒ D = 81-56

⇒ D = 25

Hence the given equation has two real roots

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{9 \pm \sqrt{25}}{2(2)}\)

⇒ \(x=\frac{9 \pm 5}{4}\)

⇒ \(x=\frac{9+5}{4} \text { or } \frac{9-5}{4}\)

⇒ \(x=\frac{14}{4} \text { or } \frac{4}{4}\)

⇒ \(x=\frac{7}{2} \text { or } 2\)

⇒ \(x=\frac{7}{2} \text { or } 2\)are the real roots.

Question 26. 5x²-9x+17=0

Solution:

Given equation is 5x²-9x+17=0

on Comparing that ax² + bx + c =0, we get

a=5, b=-19, C=17

Discriminant D=b²-4ac

⇒ 3D = (19)² – 4(5)(17)

⇒ D= 361-340

⇒ D = 21

Hence the given equation has two real roots

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{19 \pm \sqrt{21}}{2(5)}\)

⇒ \(x=\frac{19+\sqrt{21}}{10} \text { or } \frac{19-\sqrt{21}}{10}\)

⇒ \(x=\frac{19+\sqrt{21}}{10} \text { or } \frac{19-\sqrt{21}}{10} \)are real roots.

Question 27. x²-18x+77=0

Solution:

Given equation is x2-18x+77=0

On Comparing that ax²+ bx + C=0, we get

a=1, b=-18, c=77

Discriminant D= b² – 4ac

⇒ D = (-18)² -4(1)(77)

⇒ D = 394 308

⇒ D = 3·16

Hence the given equation has two real roots

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{18 \pm \sqrt{16}}{2(1)}\)

⇒ \(x=\frac{18 \pm 4}{2}\)

⇒ \(x=\frac{18+4}{2} \text { or } \frac{18-4}{2}\)

⇒ \(x=\frac{22}{2} \text { or } \frac{14}{2}\)

x = 11or7

x= 11,7 are two real numbers.

Question 28. \(\frac{2}{3} x=-\frac{1}{6} x^2-\frac{1}{3}\)

Solution:

Given equation is \(\frac{2}{3} x=-\frac{1}{6} x^2-\frac{1}{3}\)=0

On comparing that at ax2+bx+C=0, we get

⇒ \(a=\frac{1}{6}, b=\frac{2}{3}, c=\frac{1}{3}\)

⇒ \(\text { Discriminant } D=b^2-4 a c\)

⇒ \(\Rightarrow D=\left(\frac{2}{3}\right)^2-4\left(\frac{1}{6}\right)\left(\frac{1}{3}\right)\)

⇒ \(D=\frac{4}{9}-\frac{4}{18}\)

⇒ \(D=\frac{8-4}{18}\)

⇒ \(D=\frac{4}{18} \Rightarrow D=\frac{2}{9}\)

Hence the given equation has two real roots.

⇒ \(x=\frac{-b \pm \sqrt{0}}{2 a}\)

⇒ \(x=\frac{-\frac{2}{3} \pm \sqrt{\frac{2}{9}}}{2\left(\frac{1}{6}\right)}\)

⇒ \(x=\frac{\frac{-2}{3} \pm \frac{\sqrt{2}}{3}}{\frac{2}{6}}\)

⇒ \(x=\frac{\frac{-2}{3} \pm \frac{\sqrt{2}}{3}}{1 / 3}\)

⇒ \(x=\frac{(-2 \pm \sqrt{2}) \times 3}{\not 2}\)

⇒ \(x=-2 \pm \sqrt{2}\)

⇒ \(x=-2+\sqrt{2} \text { or }-2-\sqrt{2}\)

⇒ \(x=-2+\sqrt{2},-2-\sqrt{2}\) are two real roots.

Question 29. \(\frac{1}{15} x^2+\frac{5}{3}=\frac{2}{3} x\)

Solution:

Given equation is \(\frac{1}{15} x^2-\frac{2}{3} x+\frac{5}{3}=0\)

⇒ \(a=\frac{1}{15}, b=\frac{-2}{3}, c=\frac{5}{3}\)

Discriminant \(D=b^2-4ac\)

⇒ \(D=\left(\frac{-2}{3}\right)^2-4\left(\frac{1}{15}\right)\left(\frac{5}{3}\right)\)

⇒ \(D=\frac{4}{9}-\frac{20}{45}\)

⇒ \(D=\frac{20-20}{45}\)

⇒ D = 0

Hence the given equation is two real roots.

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{\frac{2}{3} \pm 0}{2\left(\frac{1}{15}\right)}\)

⇒ \(x=\frac{x}{3} \times \frac{15^{5}}{x}\)

x = 5,5 are two real roots.

Question 30. \(\sqrt{6} x^2-4 x-2 \sqrt{6}=0\)

Solution:

Given equation is √6x=4x²-2√6=0

On Comparing that ax² + bx +c=0, we get

Discriminant D= b² – 4ac

⇒ D= (4)² – 4(√c)(-2√2)

⇒ D = 16+48

⇒ D = 64

Hence Given equation has two real roots.

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{4 \pm \sqrt{64}}{2 \sqrt{6}}\)

⇒ \(x=\frac{4 \pm 8}{2 \sqrt{6}}\)

⇒ \(x=\frac{4+8}{2 \sqrt{6}} \text { or } \frac{4-8}{2 \sqrt{6}}\)

⇒ \(x=\frac{12}{2 \sqrt{6}} \text { or } \frac{-4}{2 \sqrt{6}}\)

⇒ \(x=\frac{\not 2 \times 6}{\not 2 \sqrt{6}} \text { or } \frac{-\not 2 \times 2}{\not 2 \sqrt{6}}\)

⇒ \(x=\sqrt{6} \text { or } \frac{-2}{\sqrt{6}}\)

⇒ \(\frac{-2}{\sqrt{2 \times 3}}\)

⇒ \(\frac{-6 \times 2}{2 \sqrt{6}}\)

⇒ \(\frac{-\sqrt{6}}{3}\)

⇒ \(x=\sqrt{6},-\frac{\sqrt{6}}{3}\) are two real roots.

Question 31. 256 x ² – 32x + 1 = 0

Solution:

Given equation is 256 x² = 32x+1=0

On Comparing that an’ + bx + c = 0, we get

a=256, b=-32, C=1

Discriminant D=6=4ac

⇒ D= (32) = 4(256)(1)

⇒ D= 1024-1024

⇒ D= 0

Hence given equation is two real roots

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{32 \pm \sqrt{6}}{2(256)}\)

⇒ \(x=\frac{32}{512}\)

⇒ \(x=\frac{1}{16}\)

⇒ \(x=\frac{1}{16}\) are two real roots.

Question 32. (2x+3)(3x-2)+2=0

Solution:

Given equation is (2x+3)(3x-2)+2=0

6x²-4x+9x-6+2=0

6x²+5x-4=0

On Comparing that ax²+bx+C=0, we get

a=6, b=5, C=-4

Discriminant D= b2 – 4ac

⇒ D=(C)²=4(6)(-4)

⇒ D= 25+96

⇒ D = 121

Hence given equation has two real roots

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{-5 \pm \sqrt{127}}{2(6)}\)

⇒ \(x=\frac{-5 \pm \sqrt{121}}{12}\)

⇒ \(x=\frac{-5+11}{12} \text { or } \frac{-5-11}{12}\)

⇒ \(x=\frac{6}{12} \text { or }-\frac{16}{12}\)

⇒ \(x=\frac{1}{2} \text { or }-\frac{4}{3}\)

⇒ \(x=\frac{1}{2} \text { or }-\frac{4}{3}\) are two real roots.

Question 33. x²-16=0

Solution:

Given equation is x²-16=0

x²-4²=0

(x-4)²=0

x²+4-4x=0

On comparing that ax²+6x+C=0, we get

a=1, 6=-4, c=4

Discriminant D=b²-4ac

⇒ D=(-4)2-4(1)(4)

⇒ D = 16-16

⇒ D = 0

Hence the given equation has two real roots

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{4 \pm \sqrt{0}}{2}\)

⇒ \(x=\frac{4}{2}\)

x = 2

Question 34. 36x² – 129x+(a²-b²)=0

Solution:

Given equation is 36x – 12ax + (a² – b)=0

On comparing that ax²+bx+c=0, we get

a=36, b=-12a, C=(a²– b²)

Discrimanant =) D= b²-4ac

⇒ D = (12a)²-4(36)(a²-6-b²)

⇒ D = 144a² – 144(a²-b²)

⇒ D = 144a² – 144a²+144b²

⇒ D = 144b²

Hence the given equation has two real roots

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{12 a \pm \sqrt{144 b^2}}{2(36)}\)

⇒ \(x=\frac{12 a \pm\not \sqrt{(12 b)\not^2}}{72}\)

⇒ \(x=\frac{12 a \pm 12 b}{72}\)

⇒ \(x=\frac{12(a \pm b)}{72}\)

⇒ \(x=\frac{12(a+b)}{72} \text { or } \frac{12(a-b)}{72}\)

⇒ \(x=\frac{a+b}{6} \text { or } \frac{a-b}{6}\)

⇒ \(x=\frac{a+b}{6} \text { or } \frac{a-b}{6}\) are two real roots.

Question 35. P² x² + (p²– q²)x-q²=0

Solution:

Given equation is p²x² + (p² -q²)x-q²=0

On Comparing that ax²+ bx + c = 0, we get

a=p², b= (p2 q²), c = -q²

Discriminant D= b²-4ac

⇒ D = (p2 q2)2 – 4(p2) (−22)

⇒ D= p² q⁴ + 4p² q² – 2p²q²

⇒ D = p² + 2 p² q ² + q ² =) (P²+q²) ²

Hence the given equation has two real roots.

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{-\left(p^2-q^2\right) \pm \sqrt{p^2+q p^2 q^2+q^2}}{2 p^2}\)

⇒ \(x=\frac{-p^2+q^2 \pm \not\sqrt{\left(p^2+q^2\right)\not ^2}}{2 p^2}\)

⇒ \(x=\frac{-p^2+q^2 \pm\left(p^2+q^2\right)}{2 p^2}\)

⇒ \(x=\frac{\not -p^2+q^2+\not p^2+q^2}{2 p^2} \text { or }-\frac{-p^2+\not q^2-p^2-\not q^2}{2 p^2}\)

⇒ \(x=\frac{2 q^2}{2 p^2} \quad \text { or } \quad \frac{-2 p^2}{2 p^2}\)

⇒ \(x=\frac{q^2}{p^2} \quad \text { or }-1\)

⇒ \(x=\frac{q^2}{p^2} \quad \text { or }-1\) are two real roots.

Question 36. abx² + (b²-ac)x – bc=0

solution:

The given equation is abx²+ (b²-ac)x-bc=0

on Comparing that ax² + bx + c=0, we get

a= ab, b= (b²-ac), c=-bc

Discriminant =) D= b² – 4ac

⇒ D= (b²-ac)² +4(ab) (-bc)

⇒ D= b⁴ + a²c² = 2b²ac+4ab²c

⇒ D= b⁴+ a²c²+2ab²c

⇒ 0= (b²+ac)²

Hence given equation has two real roots

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{-\left(b^2-a c\right) \pm \sqrt{\left(b^2+a c\right)^2}}{2 a b}\)

⇒ \(x=\frac{\not -b^2+a c+\not b^2+a c}{2 a b} \text { or } x=\frac{-b^2+\not a c-b^2-\not \ a c}{2 a b}\)

⇒ \(x=\frac{2 a c}{2 a b} \quad \text { or } x=\frac{-2 b^2}{2 a b}\)

⇒ \(x=\frac{c}{b} \quad \text { or } x=\frac{-b}{a}\)

⇒ \(x=\frac{c}{b} \quad \text { and} x=\frac{-b}{a}\) are two roots.

Question 37. 12abx² – (9a²-8b²) x-6ab=0

Solution:

Given equation is 12abx – (9a²-8b²)x-6ab=0

on comparing that ax2+ bx + c = 0, we get

a=12ab, b=-(9a² =8b²), c=-6ab

Discriminant ⇒ D= b²-4ac

⇒ D= (-(9a² = 8b²)² – 4(12ab) (-6ab)

⇒ D=81a⁴64b⁴-144a²b²+288a²b²

⇒ D= 81a⁴+64b⁴+ 144a²b²

⇒ D= (9a² +8b²)²

Hence the given equation has two roots

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{\left(9 a^2-8 b^2\right) \pm \sqrt{\left(9 a^2+8 b^2\right)^2}}{24 a b}\)

⇒ \(x=\frac{9 a^2-8 b^2 \pm\left(9 a^2+8 b^2\right)}{24 a b}\)

⇒ \(x=\frac{9 a^2-8 b^2+9 a^2+8 b^2}{24 a b}\) Or \(x=\frac{9 a^2-8 b^2-9 a^2-8 b^2}{24 a b}\)

⇒ \(x=\frac{18 a^x}{24 a b} \quad \text { or } x=\frac{-16 b^x}{24 a b}\)

⇒ \(x=\frac{3 a}{4 b}\) or \(x=\frac{-2 b}{3 a}\)

⇒ \(x=\frac{3 a}{4 b},-\frac{2 b}{3 a}\) are two real roots.

Question 38. Determine the nature of the roots of the following quadratic equations: 2x² + 5x – 4 = 0

Solution:

The given equation is 2x²+5x-4=0

9x²-6x+1=0

Comparing with ax + bx+c=0

a=2, 6:5, C=-4

Discriminant ⇒ D= b²– 4ac

⇒ D= (5)=4(2)(-4)

⇒ D=25+32

⇒ D= 47

⇒ D>0

Hence the equation has real and distinct roots.

Question 39. 9x²-6x+1=0

Solution:

The given equation is 9x²-6x+1=0

Comparing with ax² + bx +C=0

a=9, 6=-6, C=1

Discriminant ⇒ D= b²-4ac

⇒ D= (-6)²-(9)(1)

⇒ D= 36-36

⇒ D= 0

Hence the given equation has real and equal roots.

Question 40. Find the Value of k for which the equation 12×2+4kx+3=0 has real and equal roots.

Solution:

Given equation is 12x² + 4kx+3=0

Comparing with ax²+ bx+c=0,

a=12, b=4k, c=3

For real and equal roots,

Discriminant (D)=0 ⇒ D= b²-4ac=0

⇒(4K)²=-4(12)(3)=0

⇒ 16K²-144=0

⇒ 16k²=144

⇒ k²=\(\frac{144}{16}\)

⇒ K² = 9

⇒ k = √9

⇒ k= ±3

Question 41. Find the value of k for which the equation 2x²+5x-k=0 has real roots.

Solution:

Given equation is 2x²+5x-k=0

Comparing with ax²+bx+C=0

a=2, b=5, C=-k

For real roots, Discriminant (D) =20

(5)²+4(2)(18)20

⇒ \(k\frac{-25}{8}b^2-4 a c \geq 0\)

Question 42. The sum of a number and its reciprocal is \(\frac{10}{3}\), find the number (5).

Solution:

let the number be

According to the given Statement \(x+\frac{1}{x}=\frac{10}{3}\)

⇒ 3x²+3=10x

⇒ 3x² – 10x+3=0

⇒ 3x²-9x+x+3=0

⇒ 3x(x-3)-(x-3)=0

⇒ (3x-1)(x-3) = 0

when 3x-1=0, x = = = =

and when x-3=0, x = 3

Hence the number of (5) are 3 and 1.

⇒ \(\frac{x^2+1}{x}=\frac{10}{3}\)

⇒ \(3 x-1=0, x=\frac{1}{3}\)

and when x-3=0 , x=3

⇒ latex]3 x^2+3=10 x/latex]

Hence the numbers of are \(\frac{1}{3}\)

Linear Equation In Two Variables

Question 1. Solve for x and y:

7+y=17,

7-4=1

Solution: Given

x+y=17  → (1)

x-4=1    → (2)

From equation 1 we get y= 17-x → R

Substituting the value of from equation in 2, we get

x-(17-x)=1

2-17+2=1

2x=1+17

2x = 18

x = \(\frac{18}{2}\) ⇒ x=9

Read and Learn More Class 10 Maths

Substituting the value of x in an equation, we

y= 17-9

y= 8

Solution is x=9

y=8

2. x+2y=19

-x+2y=1

Solution: Given x+2y= 190 → (1)

-x+2y=1 → (2)

From equation 1 we get x=19-2y → 3

Substituting the value of and from equations (3)and (2) we get

-(19-2y) +2y= 1

– 19+2y+24=1

4y = 1+19

4y = 20

y = \(\frac{20}{4}\)

y = 5

Substitute value of y in equation 3 we get

x=19-25

X=19-2(5)

1=19-10

x=9

Solution is x =9

y=5

3. x – y= 0.9

\(\frac{11}{2(x+y)}=1\)

Solution: Given x – y= 0.9 → (1)

⇒ \(\frac{11}{2(x+y)}=1\)

⇒ \(\frac{11}{2 x+2 y}=1\)

2x + 2y = 11 → (2)

From equation we get x = 0-9 + y → (3)

Substituting the value of x from the equation, we get

2(0.9+4)+2y=11

1.8 +25+24=11

4y = 11-1-1

4y= 9, 2

y = \(\frac{9.2}{4} \Rightarrow y=2.3\)

Substituting the value of the y value in the (3) equation we get

2=0·9+2·3

X = 3.2

Solution is 2=3.2

y=2-3

4. 3x-2y=6

\(\frac{x}{3}-\frac{y}{6}=\frac{1}{2}\)

Solution: Given 3x-2y=6

⇒ \(\frac{x}{3}-\frac{y}{6}=\frac{1}{2}\)

⇒ \(\frac{6 x-y}{6}=\frac{1}{2}\)

2(6x-4)=6

12x-2y=6

From equation 1 we get 3x-2y=6

3x=6+24

⇒ \(x=\frac{2 y+6}{3} \rightarrow \text { (3) }\)

Substituting the value of x from equation 3 in 2, we get

⇒ \(12\left(\frac{2 y+6}{3}\right)-2 y=6\)

8y+824-2y=6

6y=6-24

6y=-18

⇒ \(y=\frac{-18}{6} \Rightarrow y=-3\)

Substituting the Value of y in Equation 3

⇒ \(x=\frac{2(-3)+6}{3}\)

⇒ \(x=\frac{-6+6}{3} \Rightarrow x=0\)

Solution is x=0

y=-3

5. 0.4x+0.34 = 1.7

07x-0-2y=0·8

Solution: Given 0.4x+0-3y=1-7 → (1)

0.72-0·24=0·8 → (2)

From equation (1) we get 0.4x=1.7-0.34

⇒ \(x=\frac{1.7-0.3 y}{0.4} \rightarrow \text { (3) }\)

Substituting the value of x from equation (3) in (2)we get

⇒ \(0.7\left(\frac{1.7-0.3 y}{0.4}\right)-0.2 y=0.8\)

⇒ \(\frac{1.19}{0.4}-\frac{0.21 y}{0.4}-0.2 y=0.8\)

2.975-0.525y-0-2y=0.8

-0-725y = 0. = 0 .525y-2.975

– 0.725y = 2175

⇒ \(0.7\left(\frac{1.7-0.3 y}{0.4}\right)-0.2 y=0.8\)

⇒ \(\frac{1.19}{0.4}-\frac{0.21 y}{0.4}-0.2 y=0.8\)

2.975y – 0.525y – 0.2y = 0.8

-0.725y  = 0.8 – 2.975

-0.725y  = -2.175

\(y=\frac{-2.175}{-0.725}\)

Substituting the value of ‘y’ from the equation

⇒ \(x=\frac{1.7-0.3(3)}{0.4}\)

⇒ \(x=\frac{1.7-0.9}{0.4}\)

⇒ \(x=\frac{0.8}{0.4}\)

x = 2

The solution is

y =3

x =2

6. y = 2x – 6

y= 0

Solution:

Given equations are

y = 2x-6 (1)

y= 0 (2)

Substituting y value in Equation (1)

0=2x-6

-2x = -6

⇒ \(x=\frac{6}{2} \Rightarrow x=3\)

Solution is x = 3

y = 0

7. 65x -33y = 97

33x-65y = 1

Solution: Given equations are 65x-33y=97 (1)

33x-65y = 10 (2)

on adding equation ( & we get

⇒ 98x-98y=98 x-y=1 → (3)

On Subtracting equation (2)from (1) we get

– 32x-32y=-96

– 32(x+y)=+96

(x + Y) = \(\frac{96}{32}\)

x + y = 3 → (4)

Adding equation (3) and (4)

putting x = 2 in equation (3)

2-y=1 =) -y= 1-2 =) -y=-1 =) y=1

Hence, the Solution is x=2

4 = 1

8. 217x+13ly=913

131x+217y=827

Solution: Given equations 217x+131y=9130 (1)

131x +2174=827 (2)

on adding (1) and (2) we get

348x+3484 = 1740

348(x+y)=1740

⇒ \(x+y=\frac{1740}{348}\) → (3)

On Subtracting the equation we get

-86x+86y=-86

-86(x-4)=-86

x-y=1 → (4)

Adding equation (3) and (4)

2x = 6

⇒ \(x=\frac{6}{2} \Rightarrow x=3\)

putting x = 3 in equation (3)

3 + y = 5

y = 5 – 3

y = 2

Hence the solution is x = 3

y = 2

Question 2. Solve the following System of equations by using the cross-multiplication method:

1. 2x+y=5

3x+2y=8

Solution: The given equation can be written as

2x+y-5=0

3x+2y-8=0

By Gross multiplication method, we get

x y 1

1 -5 2 1

2 -8 3 1

⇒ \(\frac{x}{-8+10}=\frac{y}{-15+16}=\frac{1}{4-3}\)

⇒ \(\frac{x}{2}=\frac{y}{1}=\frac{1}{1}\)

⇒ \(\frac{x}{2}=\frac{1}{1} \text { or } x=2\)

and \(\frac{y}{1}=\frac{1}{1} \text { or } y=1\)

Hence, x=2 and y=1 is the required Solution.

2.  8x+13y-29=0

12x-74-17=0

Solution:

The given equations are

8x+13y-29=0

12x-7y-17=0

By Cross multiplication method, we get

x    y   1

13 -29 8 13

-7 -17 12 -7

⇒ \(\frac{x}{-221-203}=\frac{y}{-348+136}=\frac{1}{-56-156}\)

⇒ \(\frac{x}{-424}=\frac{y}{-212}=\frac{1}{-212}\)

⇒ \(\frac{x}{-424}=\frac{1}{-212}\)

⇒ \(x=\frac{-424}{-212}\)

x = 2

and \(\frac{y}{-212}=\frac{1}{-212}\)

y = 1

Hence, x=2 and y=1 is the required Solution.

3. 2y+2x=0

4y+3x=5

Solution: The given equation Can be written as 2x +3y=0 3x+4y-5

By Cross multiplication method, we get

x    y    1

3   0    2    3

4  -5   3    4

⇒ \(\frac{x}{-15-0}=\frac{y}{0+10}=\frac{1}{8-9}\)

⇒ \(\frac{x}{-15}=\frac{y}{10}=\frac{1}{-1}\)

when \(\frac{x}{-15}=\frac{1}{-1} \Rightarrow x=15\)

and\(\frac{y}{10}=\frac{1}{-1} \Rightarrow y=-10\)

Hence, x=15 and y=-10 is the required solution.

4. \(\frac{x}{6}+\frac{4}{15}=4\)

\(\frac{x}{3}-\frac{4}{12}=\frac{19}{4}\)

Solution: The given equation can be written as

⇒ \(\frac{x}{6}+\frac{y}{15}-4=0\)

⇒ \(\frac{5 x+2 y-120}{30}=0\)

5 x+2 y-120 = 0  → (1)

⇒ \(\frac{x}{3}-\frac{y}{12}-\frac{19}{4}=0\)

⇒ \(\frac{8 x-2 y-114}{24}=0\)

8x – 2y – 114 = 0→ (2)

By Cross multiplication method & we get

x         y            1

2    – 120     5       2

-2  -114      8        2

⇒ \(\frac{x}{-225-240}=\frac{y}{-960+570}=\frac{1}{-10-16}\)

⇒ \(\frac{x}{-468}=-\frac{y}{-390}=\frac{1}{-26}\)

when \(\frac{x}{-4 6 8}=\frac{1}{-26}\)

⇒ \(x=\frac{-468}{-26} \Rightarrow x=18\)

⇒ \(\text { and } \frac{y y}{-390}=\frac{1}{-26} \Rightarrow y=\frac{-390}{- 20} \Rightarrow y=15\)

Hence X-18 and y = 15 is the required solution.

5. x+y=a+b

ax-by=a2=62

Solution: The given equations Can be written as

x+y=(a-6)=0

ax-by-(a2+63)=0

By Cross multiplication method, we get

x            y               1

1      -(a-b)        1       1

-b     -(a2+b2)   a    -b

⇒ \(\frac{x}{-\left(a^2+b^2\right)-b(a-b)}=\frac{y}{-a(a-b)+\left(a^2+b^2\right)}=\frac{1}{-b-a}\)

⇒ \(\frac{x}{-a^2-b^2-a b+b^2}=\frac{y}{-a^2+a b+a^2+b^2}=\frac{1}{-b-a}\)

⇒ \(\frac{x}{-a^2-a b}=\frac{y}{a b+b^2}=\frac{1}{-b-a}\)

when \(\frac{x}{-a^2-a b}=\frac{1}{-b-a}\)

⇒ \(x=\frac{-b(a+b)}{-(a+b)}\)

x = a

and \(\frac{y}{a b+b^2}= \frac{1}{-b-a}\)

⇒ \(\frac{y}{-b(-a-b)}=\frac{1}{-b-a}\)

⇒ \(y=\frac{-b(-b-a)}{+b-a}\)

Hence x=a and y = b are the required Solution

6. ax-by=a²+b²

x+y=20

Solution: The given equations Can be written as,

ax-by-(a²=+b²)=0

x+y-2a=0

By Cross multiplication method, we get

x                y             1

-b         -(a²+b²)      a      -b

1               -2a           1        1

⇒ \(\frac{x}{2 a b+\left(a^2+b^2\right)}=\frac{y}{-\left(a^2+b^2\right)+2 a^2}=\frac{1}{a+b}\)

⇒ \(\frac{x}{2 a b+\left(a^2+b^2\right)}=\frac{y}{-\left(a^2+b^2\right)+2 a^2}=\frac{1}{a+b}\)

⇒ \(\frac{x}{(a+b)^2}=\frac{y}{a^2-b^2}=\frac{1}{a+b}\)

⇒ \(\frac{x}{(a+b)^2}=\frac{1}{a+b}\)

⇒ \(x=\frac{(a+b)^2}{(a+b)}\)

x = a+b

and \(\frac{y^2}{a^2-b^2}=\frac{1}{a+b}\)

⇒ \(y=\frac{(a+b)(a-b)}{(a+b)}\)

y = a-b

Hence x= a+b an is the required solution.

7. ax+by = c

bx+ay=1+c

Solution: The given equations can be written as

ax+by = c

bx+ay= (C+1)=0

B Cross multiplication method, we get

x                  y                  1

b                -c                 a             b

a             -(c+1)            b               a

⇒ \(\Rightarrow \frac{x}{-b(c+1)+a c}=\frac{y}{-b c+a(c+1)}=\frac{1}{a^2-b^2}\)

⇒ \(\frac{x}{-b c-b+a c}=\frac{y}{-b c+a c+a}=\frac{1}{a^2-b^2}\)

⇒ \(x=\frac{ac-b-b c}{a^2-b^2}\)

and \(\frac{y}{-b c+a c+a}=\frac{1}{a^2-b^2}\)

⇒ \(y=\frac{-(b c-a-a c)}{-\left(-b^2-a^2\right)} \Rightarrow y=\frac{b c-a-a c}{b^2-a^2}\)

Hence x \(=\frac{a c-b-b c}{a^2-b^2} \text { and } y=\frac{b c-a-a c}{b^2-a^2}\).

8.  (a-b)x + (a+b)y = a2- 2ab-b2

(a+b) (x+y)= a2+b2

Solution:

The given equation can be written as

(a-b)x + (a+b)y – (a2+2a+b2) = 0

(a-b)x + (a+b)y – (a+b)2 = 0 (1)

(a+b)(x+y) = a2+b2

ax+ay+bx+by – (a2-b2) = 0

(a+b)x + (a+b)y – (a-b)2 = 0 (2)

By cross-multiplication methods (1)and(2)

x              y                 1

(a+b)      -(a+b)2      (a-b)     (a+b)

(a+b)      -(a2-b2)     (a+b)    (a+b)

⇒ \(\frac{x}{-(a+b)\left(a^2-b^2\right)+(a+b)^3}=\frac{y}{-(a+b)^3+\left(a^2-b^2\right)(a-b)}=\frac{1}{(a-b)(a+b)-(a+b)^2}\)

⇒ \(\frac{x}{-a^3+a b^2-a^2 b+b^3+a^3+b^3+3 a^2 b+3 a b^2}\)

⇒ \(=\frac{y}{-p^2-p^2-3 a^2 b-3 a b^2+a^3-a^2 b-a b^2+b^3}\)

⇒ \(=\frac{1}{a\not^2-b^2-\not a^2-b^2-2 a b}\)

⇒ \(\frac{x}{2 b^3+4 a b^2+2 a^2 b}=\frac{y}{-4 a^2 b-4 a b^2}=\frac{1}{-2 b^2-2 a b}\)

⇒ \(\frac{x}{2\left(b^3+2 a b^2+a^2 b\right)}=\frac{y}{-4\left(a^2 b-a b^2\right)}=\frac{1}{-2 b(b+a)}\)

when \(\frac{x}{2\left(b^3+2 a b^2+1 a^2 b\right)}=\frac{1}{-2 b(b+a)}\)

⇒ \(x=\frac{2\not\left(b^3+2 a b^2+a^2 b\right)}{-2 b(b+a)}\)

⇒ \(x=\frac{\not b\left(b^2+2 a b+a^2\right)}{\not b(b+a)}\)

⇒ \(x=\frac{(a+b)^2}{(a+b)}\)

x = a+b

and\(\frac{y}{-4\left(a^2 b-a b^2\right)}=\frac{1}{-2 b(b+a)}\)

⇒ \(y=\frac{-4 b\left(d^2-a b\right)}{-2 b(b+a)}\)

⇒ \(y=\frac{2 a^2-2 a b}{a+b} \times \frac{1}{2 a^2}\)

⇒ \(y=\frac{-2 a b}{a+b}\)

Hence x=a+b and \(y=\frac{-2 a b}{a+b}\) is the required solution.

Question 3. Each of the following System of equations determines whether the System has a unique solution, no Solution, or infinitely many Solutions. In Case there is a unique solution, find it:

1. 2x-3y=17

4x+9=13

Solution: Given equations are

2x-3y=170 (1)

4x+y=13 (2)

Here a₁ =2, b₁ =-3, C₁ = 17

a₂ =4, b₂ =1 and C₂ =13

Now, \(\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2} \text { and } \frac{b_1}{b_2}=\frac{-3}{1}\)

Since\(\frac{a_1}{a_2} \neq \frac{b_1}{b_2} .\) Hence, the given system has a unique solution.

By Cross multiplication method, we have

x              y          1

3      -17         2         -3

1      -13        4           1

⇒ \(\frac{x}{39+17}=\frac{y}{-68+26}=\frac{1}{2+12}\)

⇒ \(\frac{x}{56}=\frac{y}{-42}=\frac{1}{14}\)

When \(\frac{x}{56}=\frac{1}{14}\)

⇒ \(x=\frac{56}{14} \Rightarrow x=4\)

And \(\frac{y}{-42}=\frac{1}{14}\)

⇒ \(y=\frac{-42}{14} \Rightarrow y=-3\)

Hence x=4 and y=-3 is the required solution.

2. 5x+2y=16

3x+ \(\frac{6}{5}\) y = 2

Solution: Given equation are 5x+2y=16

3x+\(\frac{6}{5}\)

Here a₁=5, b₁=2, c₁=16

a₂=3, b₂ = \(\frac{6}{5}\), c₂=2

Now,

⇒ \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)

⇒ \(\frac{5}{3}=\frac{2}{6 / 5} \neq \frac{16}{2}\)

⇒ \(\frac{5}{3}=\frac{10}{6} \neq 8\)

⇒ \(\frac{5}{3}=\frac{5}{3} \neq 8\)

Since,\(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2} .\) hence the given system has no solution.

3. 3x+4=2

6x+2y=4

Solution: Given equations are 3x+y=2

6x+2y=4

Here a₁ = 3, b₁ = 1, c₁=2

a₂=6, b₂=2, c₂=4

Now,

⇒ \(\frac{a_1}{a_2}=\frac{3}{6} \text { and } \frac{b_1}{b_2}=\frac{1}{2} \text { and } \frac{c_1}{c_2}=\frac{2}{4}\)

Since \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} .\) Hence the given system has an infinite solution.

4. \(\frac{x}{3}+\frac{y}{2}=3\)

x-2y = 2

Solution: Given equations are \(\frac{x}{3}+\frac{y}{2}=3\)

x-2y = 2

Here a₁ = \(\frac{1}{3}\), b₁ = \(=\frac{1}{2}\)
, c₁=2

a₂=6, b₂=2, c₂=4

Now

⇒ \(\frac{a_1}{a_2}=\frac{1 / 3}{1}=\frac{1}{3} \text { and } \frac{b_1}{b_2}=\frac{1 / 2}{-2}=\frac{-1}{4}\)

Since \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2} .\) Here the given system has unique solutions

By Cross multiplication method, we get

x           y             1

1/2        -3          1/3        1/2

-2          -2          1           -0

⇒ \(\frac{x}{-\left(\frac{1}{2}\right)(2)-6}=\frac{y}{-3+2\left(\frac{1}{3}\right)}=\frac{1}{-2\left(\frac{1}{3}\right)-1\left(\frac{1}{2}\right)}\)

⇒ \(\frac{x}{-1-6}=\frac{y}{-3+\frac{2}{3}}=\frac{1}{-\frac{2}{3}-\frac{1}{2}}\)

⇒ \(\frac{x}{-7}=\frac{y}{\frac{-9+2}{3}}=\frac{y}{\frac{-4-3}{6}}\)

⇒ \(\frac{x}{-7}=\frac{y}{-7 / 3}=\frac{1}{-7 / 6}\)

⇒ \(\frac{x}{-7}=\frac{-3 y}{7}=\frac{-6}{7}\)

⇒ \(\frac{x}{-7}=\frac{-3 y}{7}=\frac{-6}{7}\)

⇒ \(\text { when } \frac{x}{-7}=\frac{-6}{7}\)

⇒ \(x=\frac{-6(-7)}{7}\)

⇒ \(x=\frac{42}{7}\)

x = 6

and \(\frac{-3 y}{7}=\frac{-6}{7}\)

⇒ \(-3 y=\frac{-6(7)}{7}\)

7(-3)=-6(7)

-21y = -42y

⇒ \(y=\frac{42}{21}\)

y = 2

Hence x = 6 and y=2 is a required solution.

5. Kx+2y=5

3x+9=1

Solution: The given system of equations is

kx+2y=5

3x+y=1

Here, a₁=k, b₁ =2 and c₁=5

a₂=3, b₂=1 and c₂=1

The System has unique Solution \(\text { if } \frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

⇒ \(\frac{k}{3} \neq \frac{2}{1}\)

⇒ \(\Rightarrow \quad k \neq 6\)

So, k can take any real value except 6.

6. 4x+ky +1=0

2x+2y+2=0

Solution: The given System of equations is 4x+ky+8=0

2x+2y+2=0

Here a₁ = 4, b₁ = k and c₁ =8

a₂=2, b₂ =2 and C₂ =2

The System has unique solution if \(\text { if } \frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

⇒ \(\begin{aligned}
& \Rightarrow \frac{2 y}{x} \neq \frac{k}{2} \\
& \Rightarrow k \neq 4
\end{aligned}\)

So k can take any real value except 4.

7. 2x-37-5=0

kx-by-8=0

Solution: The given System of equations are 2x-34-5=0

kx-6x-8=0

Here, a₁ =2, b₁=3 and c₁ =-5

a₂=k, b₂=6 and C₂ =-8

The System has unique solution if \(\text { if } \frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

⇒ \(\frac{2}{k}+\frac{-3}{-6}\)

⇒ \(\frac{2}{k}+\frac{1}{2}\)

⇒ \(\begin{aligned}
& \frac{4}{k} \neq 1 \\
&\quad k \neq 4
\end{aligned}\)

So k can take any real value except 4.

8. 8x+5y=9

kx+10y=18

Solution: The given System of equations is 8x+5y=9

Here a₁ =8, b₁=5 and c₁ =9

a₂ =k, b₂ = 10 and c₂ = 18

kx+10y=18

The System has infinitely many solutions if \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

⇒ \(\frac{8}{k}=\frac{5}{10}=\frac{9}{18}\)

⇒ \(\frac{8}{k}=\frac{1}{2}=\frac{1}{2}\)

when \(\frac{8}{k}=\frac{1}{2} \Rightarrow k=16\)

Hence, for k = 16, the given System equations will have infinitely many solutions.

Question 4. The Sum of the two numbers is 85. Suppose the langer number exceeds four times the Smaller one bys. Find the numbers.

Solution: let the two numbers x and y, and x>y.

According to question x+y=85 → (1)

and X-4y=5 → (2)

Subtracting equation (2) from (1), we get

5y=80 =) y=16

Putting y=16 in equation (, we get

x+16=85

2=85-16

– x = 69

Real Numbers Exercise – 1.1

Question .1 Use Euclid division algorithm, to find the H.C.F of the following:

1. 70 and 40
2. 180 and 45
3. 165 and 225
4. 155 and 1385
5. 105 and 135
6. 272 and 1032

Solution:

1. 70 and 40

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Here 40 > 70

7o = 4O * 1 + 30

4o = 30 * 1 + 10

3o = 10 * 1 + 0

Since remainder = 0

The recent divisor is the H.C.F

H.C.F (40,70)

2. 180 and 45

Hare 18 > 45

45 = 18 * 2 + 9

18 = 9 * 2 + 0

Since remainder = 0

The recent divisor is the H.C.F

H.C.F ( 18,45) = 9

3. 165 and 225

Here 165 > 225

225 = 165 * 1 + 90

165 = 90 * 1 + 75

90 = 75 * 1 + 15

75 = 15 * 5 + 0

Since remainder = 0

The recent divisor is the H.C.F

H.C.F ( 165,225) = 15

4. 155 and 1385

Hare 155 and 1385

1385 = 155 * 8 + 145

155 = 145 * 1 + 0

145 = 10 * 14 + 5

10 = 5 * 2 + 0

Since remainder = 0

The recent divisor is the H.C.F

H.C.F ( 155,1385) = 15

5. 105 and 135

Here 105 and 135

135 = 105 * 1 + 30

105 = 30 * 30 + 15

30 = 15 * 2 + 0

Since remainder = 0

The recent divisor is the H.C.F

H, C.F ( 105, 135)

6. 272 and 1032

Here 272 and 1032

1032 = 272 * 3 + 216

272 = 216 * 1 + 56

216 = 56 * 3 + 48

48 = 8 * 6 + 0

Since remainder = 0

The recent divisor is the H.C.

H, C.F ( 272, 1032)

Question: 2. The H.C.F of 408 and 1032 is expressible in the form of 1032×2-408xy, then find the Value of y.
Solution:

First, we will find the H-CF of 408 and 1032

Here 408 >1032

1032 = 408 x 2 + 216

408 = 216 x 1 + 192

216= 192 x 1 + 24

192=24 x 8 + 0

Since remainder = 0

The recent divisor is the H-C.F

H.C.F. of (408, 1032) = 24

Now, 1032×2-408xy = 24

= -408xy = 24-1032X 2

-y = \(\frac{24-2064}{408}\)

-y = \(\frac{-2040}{408}\)

y = 5

Question 3. If the H.C.F of 56 and 72 is expressible in the form of 56x+72×53, then find the Value of x.
Solution:

First, we will find the H.C.F of 56 and 72

Here 56 > 372

72 = 56 x 146

56= 16 x 3 + 8

16 = 8 x 2 + 0

Since remainder = 0

The recent divisor is the H·C.F

H.C.F. of (56,72)=8

Now, 56x+72×53 = 8

56x = 8 – 72 x 53

56x= 8-3,816

x = \(\frac{-3,808}{56}\)

x = – 68

Question 4. Express the H.C.F of 18 and 24 in the form
solution:

Here 18>24

24 = 18× 1+6

18=6×2+6

6=6×1+0

Since remainder =0

The recent divisor is the H.C.F

H.C.F. (18,24)=6

Now,

6=18-6×2

6=18-(24-18X1)

= 18-24 +18 x |

18×2-24 = 18x +244

where x = 2, y=-1

Question 5. Express the H.CF of 30 and 36 in the form of 30x + 36y.

Solution:

Here 30> 36

30= 200 6×4+6

6= 6X1+0

Since remainder = 0

The recent divisor is the H.C.F

H.C.F. (30,36) = 6

Now, 6=30-6×4

6=30-(36-30×1)

= 30-36+30X1

= 30X2-36

= 30x+364

where, x=2 and y=-1

Question 6. Find the largest number that divides 189 and 249 9 in each case.

Solution: We have to find a number, which divides the other numbers

Means H.C.F.

It is then that the required Number, when divided between 189 and 249 leaves the remainder 9; 9 is extra in each number. It means that if these numbers are 6 less, then there is no remainder in each case.

89-9=180 and 249-9=240 are completely divisible by the required number.

H.C.F. (180,240) = 60

Hence, the required number =  60.

Question 7. Find the largest number that divides 280 and 1248 u and 6 respectively. leaving the remainder solution.

Solution: We have to find a number, which divides the other numbers means → H.C.F.

It is given that the required number, when divided between 280 and 1248, leaves the remaining 4 and 6 respectively. It means that if 280 is 4 less than, 1248 is 6 less, then on division, gives no remainder.

280-4=276 and 1248-6=1242 are Completely divisible to the required number.

First, we will find the H.C.F of 276 and 1242

H.C.F. (2.76, 1242) = 138

Hence, the required number =138.

Question 8. Find the greatest number that divides 699, 572, and 442 leaving remainders 6, 5, and 1 respectively.

Solution: We have to find a number, which divides the other numbers means → H.C·F.

It is given that the required numbers when divided into 699, 572, and 442, leave the remaining 6,5 and I respectively. It means that if 699 is 6 less, 572 is s less, and 442 is I less, then on division, gives no remainder.

699-6=693, 572-5=567 and 442-1=441 are and 442-1=441 are completely by the required number .

First, we will find the H.C.F of 693 and 567.

693 = 567×1 +126

567 = 126×4 +63

126 = 63X2  + o

Now, we filled the H.C.F of 63 and 441.

441=63×7 +0

H.C.F. (63, 441) = 63

Required Number = 63

Question 9. A sweet Seller has 420 kaju burfis and 130 badam burfis. She wants to stack them in such a way that each stack has the same number, and they take up the least area of the tray what is the number of Sweets that Can be Placed In each stack for this purpose? Also, find the number of stacks.

Solution: Maximum number of burfis in each stack = H.C.F of 420 and 130

420 = 2X2 X3 X5 X 7

130 = 2X5X 13

H·CF = 2×5 = 10

Maximum number of burfis in each stack = 10

Also, number of stacks = \(=\frac{420}{10}+\frac{130}{10}=42+13=55 .\)

Question 10. Three sets of English, Hindi, and Mathematics books have to be stacked In such a way that all the books are stored topicwise and the height of each stack is the Same. The number of English books is 96, the number of Hindi books is 240 and the number of mathematics books is 336. Assuming that the books are of the Same thickness, determine the number of Stacks of English, Hindi, and Mathematics books and hence the total number of Stacks.

Solution: Maximum number of books in each stack H.C.F. of

96,240 and 336

96= 2x2x2x 2 x 2 x 3

240 = 2x2x2x2x3x5

336 = 2x2x2 × 2 × 3 × 7

H·C.F = 2x2x2x2x3

Maximum number of books in each stack = 48
Also, number of stacks = 96

Also number Stacks = \(\frac{96}{48}+\frac{240}{48}+\frac{336}{48}\)

Real Numbers Exercise 1. 2

Question 1. Express each of the following as a product of prime factors:

1. 96
2. 48
3. 150
4. 3072

Solution:

96 = 2 * 2 * 2 * 2 * 2 * 3

96 = 25 * 3

84 = 2 * 2 * 3 * 7

84 = 22 * 3 * 7

150=2 x 3 x 5 x 5

150 = 2 x 3 x 52

3072 = 2× 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

3072= 210 X 3

Question 2. Find the H.C.F and L.C.M of the following pairs using the prime factorization method:

1. 12 and 25
2. 20 and 25
3. 96 and 404
4. 336 and 56

Solution:

1. 12 and 25

Now

H.C.F = 3

and L.C.M= 2 x 2 x 3 x 5

= 60

2. 20 and 25

Now

H.C.F = 5

and L.C.M = 2 x 2 x 5 x 5

= 100

3. 96 and 404

Now

H.C.F = 2 x 2= 4

and L.C.M = 2 x 2 x 2 x 2 x 2 x 3 x 101

= 9696

4. 336 and 56

Now

H.C.F = 2 x 2 x 2 x 7 = 56

and L.c.M = 2 x 2 x 2 x 2 x 3 x 7 = 336

Question 3. Using the prime factorization Method, find the HC.F and L.C.M of the following Pairs. Hence Verify H-C.F. XL.C.M= Product of two numbers.

1. 96 and 120
2. 16 and 20
3. 396 and 1080
4. 144 and 192

Solution:

1. 96 and 120

96=2x2x2 x 2 x 2 x 3

120 = 2x 2 x 2 × 3 × 5

Now, H.C.F. = 2×2×2×3=24

and LCM2 = 2 X 2 X 2 X 2 X 2 X 3 X 5 = 480

Now H.C.F X L.C. M = 12×144=24×480 = 11,520

and product of two numbers = 96X120 = 11,520

Hence, H-C.F X L.C.M = Product of two numbers

2. 16 and 20

16 = 2 X 2 X 2 x 2

L.C.M = 2 * 2 * 2 * 5 = 80

Now H.C.F x L .C.F = 4 * 80 = 320

and product of two numbers = 16 * 20

Hence H.C.F= product of two numbers

3. 396 and 1080

396 = 2×2 × 3 × 3 × 11

1080 = 2x2x2 × 3 × 3 × 3 × 5

Now, H.C.F. = 2 x2 x3x3 =36

and L.C.M. = 2X2 X2 X3 X3X3 X5x11

=11880

Now H.C.FXL.CM = 36X11880 = 4,27,680 and Product of two numbers = 4,27,680

Hence, H.C.F. XL.C.M = Product of two numbers

4. 144 and 192

144 = 2x2x 2×2×3×3

192 = 2x2x2 × 2 × 2 × 2 × 3

Now, H.C.F=2x2x2x2 x3 = 48

and L.C.M= 2×2 X2 X2 X2 X2 X3 X3

= 576

Now H.C.FXL.CM = 48X576 = 27,648 and Product of two numbers = 144 X 192

= 27,648

Hence, H.C.F. X  L.C.M.= product of two numbers

Question 4. The H.C.F and LCM of the two numbers are 145 and 2175 respectively. If the first number is 435, find the second number.

Solution:

Here, H.GF = 145

L.C.M = 2175

Now, First no. x Second no. = H.C.F. X L.C.M.

Second no = \(\frac{\text { H.C.F. XLC.M }}{\text { Firstno. }}\)

Second no = \(\frac{145 \times 2175}{435}\)

Second no = \(\frac{315375}{435}\)

Question 5. check whether 18 ⁿ can end with the digit o for the natural number n.

Solution:

18= 2 x 3 x 3 = 22 x 32

18n = (2×32) ⁿ = 2 ⁿ x32 ⁿ

It has no term containing

No Value of MEN for which 18 ⁿ ends with digit 0.

Question 6. On a morning walk, three persons step off together and their steps are 40cm, 42cm, and us cm respectively, what is the difference minimum distance each should walk so that each Can Cover the Same distance In Complete Steps?

Solution: We have to find a number (distance) that is divided by each number Completely, which means → L.CM

we have to find the L.C.M. of 400m, 42cm, and our cm to get the required distance.

Now, L. C.M = 2 x 2 x 2 x 5×3×7 × 3 = 2520

Minimum distance each should walk = 2520 Cm

Question 7. Write the missing numbers in the following factor tree:

Solution:

1. The upper box, on 7 and 13 is filled by the Product of 7 and 13, 9.
2. The upper next box, on Sand 91 is filled by the product of 5 and 91, 1.e., 455
3. The upper next box, on 3 and USS is filled by the product of 3 and USS, i.e., 1365
4. The topmost box, on 3 and 1365 will be filled by the product of 3 and 1365 i.e, 4095

Question 8. State whether the given statements are true or false:
Solution:

1. The Sum of two rationals is always rational. (True)
2. The Sum of two irrationals is always irrational. (False)
3. The product of two rationals is always rational. (True)
4. The product of two irrationals is always irrational, (False)
5. The Sum of a rational and an irrational is always rational. (False)

The product of a rational and an irrational is always rational. (True)

CBSE Solutions For Class 10 Mathematics Chapter 2

Question 1. x²+9x+20

Solution:

Let P(x) = = x²+9x+20

= x²+5x+4x+20

= x(x+5)+4(x+5)

(x+4)(x+5)

P(x)=0

(x+4)(x+5)=0

x+4=0 or x+5=0

x=-4 x=-5

Zeros of P(x) are -4 and -5 \(=\frac{-9}{1}=\frac{- \text { Coefficient of } x}{\text { coefficient of } x^2}\)

Now, Sum of Zeros = -4+(-5) \(\frac{20}{1}=\frac{\text { Constant term }}{\text { Coefficient of } x^2}\)

Question 2. x² – 9

Solution:

Let P(x) = x² = 9 = x²=-32 = (x-3)(x+3)

P(x)=0

(x-3)(x+3)=0

x-3=0 Or x+3=0

x=3 x=-3

Zeros of p(x) are 3 and -3

Now, Sum of Zeros = 3+(-3)=0 = \(\frac{-0}{1}=-\frac{\text { Coefficient of } x}{\text { Coefficient of } x^2}\)

and Product of Zeros = (3)(-3) = −9 = \(\frac{-9}{1}=\frac{\text { Constant term }}{\text { Coefficient of } x^2}\)

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Question 3. Find the quadratic polynomials, the Sum of whose Zeroes is 17 and the product is 60. Hence, find the Zeroes of the polynomial.

Solution:

Let and B be the Zeroes of the polynomial P(X).

Given that X+B=17 and αβ=60

Now, P(x) = x²= (α+β)γ + αβ

= x²= 17x+60

= x²=12x-5x+60

= x(x-12)-5(x-12)

= (x-5)(x-12)

There may be So many different polynomials that satisfy the given Condition. The general equation quadratic polynomial will be k(x2=-17x+60), where k = 0

P(x)=0

(x-5) (x-12) = 0

(x-5)=0 or (x-2)=0

2=5 Or x=12

Zeros are 12 and 5.

Question 4. Find a quadratic polynomial, the Sum of whose Zeros is 7 and the product is -60. Hence, verify the relation between Zeros and Coefficients of the polynomial.

Solution:

Let and B be the Zeros of the polynomial P(x).

Given that α+β=ϒ and αβ = -60

Now, P(x) = x² – (α+β) γ+ αβ

= x²-7x-60

= x²-12x+5x-60

= x(x-2)+5(x-12)

= (x-12)+(x-5)

There may be so many different polynomials which satisfy the given Condition. The general quadratic polynomial will be k (x²-7x-60), where k = 0.

P(x)=0

(x-12) (x+5)=0

(x-12)=0

x=12 or (x+5)=0

x=-5

Zeros are 12 and -5.

Question 5. If the product of Zeroes of the polynomial 30+ 5x+k is 6, find the value of k.

Solution:

Given polynomial = 31751+k

Product of Zeroes = \(\frac{\text { Constant term }}{\text { Coefficient of } x^2}\)

6= \(\frac{k}{3}\)

= 6×3=k

⇒ k = 18

Question 6. If the Sum of Zeroes of the polynomial x²+2x-12 is 1, find the value of k.

Solution:

Given polynomial = x²+2kx-12

Sum of Zeroes = \(-\frac{\text { Coefficient of } x}{\text { Coefficient of } x^2}\)

1 = \(\frac{-2 k}{1}\)

1 = -2K

⇒ \(k=\frac{-1}{2}\)

Question 7. If x = \(\frac{5}{3}\) and x = \(\frac{-1}{2}\) are the Zeroes of the polynomial ax=7x+b, then find the values of a and b.

Solution: Let P(x) = ax27x+b

⇒ \(x=\frac{5}{3} \text { and } x=\frac{-1}{2}\) are zeroes of p(x)

⇒ \(P\left(\frac{5}{3}\right)=0\)

⇒ \(a\left(\frac{5}{3}\right)^2-7\left(\frac{5}{3}\right)+b\)

⇒ \(\frac{25 a}{9}-\frac{35}{3}+b\)

⇒ \(b=\frac{-25 a}{9}+\frac{35}{3}\)

⇒ \(P\left(\frac{-1}{2}\right)=a\left(\frac{-1}{2}\right)^2-7\left(\frac{-1}{2}\right)+b\)

⇒ \(\frac{a}{4}+\frac{7}{2}+b\)

⇒ \(\frac{a}{4}+\frac{7}{2}-\frac{25 a}{9}+\frac{35}{3}=0\)

⇒ \(\frac{a}{4}-\frac{25 a}{9}=-\frac{7}{2}-\frac{35}{3}\)

⇒ \(\frac{9 a-100 a}{36}=\frac{-21-70}{6}\)

⇒ \(-\frac{91 a}{36}=\frac{-91}{6}\)

⇒ \(\frac{91 a}{6}=91\)

91a = 546

⇒ \(a=\frac{546}{91}\)

a = 6

⇒ \(b=\frac{-25(6)}{9}+\frac{35}{3}\)

⇒ \(b=\frac{-25(6)}{9}+\frac{35}{3}\)

⇒ \(b=\frac{-150+105}{9}\)

⇒ \(b=\frac{-45}{9}\)

b = -5

Question 8. Verify that 1,-2, 4 are Zeros of the Cubic polynomial x³-3x²-6x+8. Also, Verify the relation between Zeroes and Coefficients of the polynomial.

Solution:

Here, P(x) = x³-3x²-6x+8

P(1) = (1)² = 3(1)² – 6(1) +8 = 1-3-6+ 8 = −9+9=0

P(-2) = (2) ³ -3(-2)=6(-2)+8=-8-12+12+8 = 0

P(4) = (4)³-3(4) -6(4) +8 = 64-48-24+8 = 0

1-2 and 4 are Zeroes of P(x).

Now, α+B+= 1-2+4=3 = \(\frac{-3}{1}=-\frac{\text { Coefficient of } x^2}{\text { Coefficient of } x^3}\)

XB+B++√α = (1)(-2)+(-2) (4)+(4)(1) = −2-8+4

= -10+4

⇒\(\frac{-6}{1}=\frac{\text { Coefficient of } x}{\text { Coefficient of } x^3}\)

and LB7 = (1)(-2) (4) =\(-\frac{8}{1}=\frac{8}{1}=\frac{\text { Constant term }}{\text { Coefficient of } x^3}\)

Question 9. Verify that 2-4. and are zeroes of the Cubic polynomial 3x ³ +5x² = 26x+8. Also, verify the relation between Zeroes and Coefficients of the polynomial.

Solution:

Here, P(x)=3x³ +5 x 2 – 26x+8

P(2) = 3(2)3 +5(2)2 = 26(2)+8=24+20-52+8 = 52-52 = 0

P(-4)=3(-4)3 + 5(-4)=26(-4)+8=-(92+80+104+8=-192+192=0

⇒ \(P\left(\frac{1}{3}\right)=3\left(\frac{1}{3}\right)^3+5\left(\frac{1}{3}\right)^2-26\left(\frac{1}{3}\right)+8=\frac{3}{27}+\frac{5}{9}-\frac{26}{3}+8=\frac{3+15-234+216}{27}\)

⇒ \(=\frac{234-234}{27}=0\)

2,-4. and — are Zeroes of P(x),

Now \(\text { , } \alpha+\beta+\gamma=2-4+\frac{1}{3}=-2+\frac{1}{3}=\frac{-6+1}{3}=-\frac{5}{3}=\frac{5}{3}=-\frac{\text { Coefficient of } x^2}{\text { Coefficient of } x^3}\)

⇒ \(\alpha \beta+\beta \gamma+\gamma \alpha=(2)(-4)+(-4)\left(\frac{1}{3}\right)+\left(\frac{1}{3}\right)(2)=-8-\frac{4}{3}+\frac{2}{3}=\frac{-28+2}{3}=\frac{-26}{3}\)

⇒ \(=\frac{- \text { Coefficient of } x^2}{\text { coefficient of } x^3}\)

and \(\alpha_\beta \beta=(2)(-4)\left(\frac{1}{3}\right)=-\frac{-8}{3}=\frac{8}{3}=\frac{\text { Constant term }}{\text { Coefficient of } x^3}\)

Question 10. Find a Cubic polynomial whose Zeroes are 5, 6, and -4.

Solution:

Let α= 5, β=6 and γ=-4

α+β+ γ = 5+6-4 =) ||- 4 = 7

αB+ βγ + γα = 5(6)+6(-4)+(-4) (5)

= 30-24-20

= 30-44

= -14

XB = 5(6)(u)

= -120

Cubic Polynomial = x³-(x+β++) x² + (αβ+ βγ + γα) x-px

= x² – (7) x2 + (-14)x-(-120)

= x³-7x² = 14x+120

Question 11. Find a Cubic polynomial whose Zeroes are 11 and -1.

Solution:

let α= \(\frac{1}{2},\),B = 1 and 2 =-1

⇒ \(\alpha+\beta+\gamma=\frac{1}{2}+1-1=\frac{1}{2}\)

⇒ \(\alpha_\beta+\beta^1+\gamma \alpha=\frac{1}{2}(1)+(1)(-1)+(-1)\left(\frac{1}{2}\right) \Rightarrow \frac{1}{2}-1-\frac{1}{2} \Rightarrow-1\)

⇒ \(\alpha \beta \gamma=\frac{1}{2}(1)(-1) \Rightarrow \frac{-1}{2}\)

Cubic polynomial = x³= (α+β+γ) x2 + (αβ+ βγ + γα)x-αßγ

⇒ \(x^3-\frac{1}{2} x^2+(-1) x-\left(-\frac{1}{3}\right)\)

⇒ \(x^3-\frac{1}{2} x^2-x+\frac{1}{2}\)

⇒ \(2 x^3-x^2-2 x+1\)

Question 12. Find the quotient and remainder in each of the following and verify the division algorithm:

1. P(x) = x²=14x²+2x-1 is divided by g(x)=x+2

Solution:

Now, quotient = x2=6x+14, remainder = -29

dividend = x³-4x²+2x-1 and divisor = x+ 2

and quotient x divisor + remainder = (x²-6x+(4)(x+2)-29

= x²-6x²+14x+2x²=1271+28-29

= x²-4x² +2x-1

= dividend

2. P(x) = x² + 2x²= x + 1 is divided by g(x) = x²+1

Solution:

Now, quotient = x²+1, remainder = -x, dividend = x² + 2x = x +1 and

divisor = x+1

and quotient divisor + remainder = (x² + 1)(x² + 1) – α = ) x + x2 + x² + 1-x

x 4 + 2x = x + 1 =) dividend

Question 13. Actual division shows that x+2 is a factor of x³ + 4x²+3x-2.

Solution:

Question 14. If I am a zero of the polynomial x²-4x²=7x+10, find its other two Zeroes

Solution:

let P(x)=x³=_472-70+10

x= 1 is a zero of P(x)

(x-1) is a factor of P(x)

P(x) = x3- 4x²-7x+ 10 =) (x-1) (x=3x-10)

(x-1) (x²+2x-5x-10)

(x-1) ((x(x+2)-5(x+2))

(x-1)(x+2)(x-5)

Now, P(x)=0

⇒ (x-1)(x+2)(x-5)=0

⇒ α-1=0 or x+2=0 or x-5=0

x=1 or x=-2 or x=5

Hence, other Zeroes are -2 and 5

Question 15. If Land -2 are two Zeroes of the polynomial x⁴+x³=11x=9x+18, find the other two Zeroes.

Solution:

Let P(x) = x² + x² – 1172=9x+18

x=1, x=-2 is a zero of p(x)

(x-1) (x+2) is a factor of P(x)

x²+2x-x-2 =) x²+x-2 is a factor of P(x)

p(x0 = x4 = x3 – 11×2 – 9x + 18 = (x-1)(x+2)(x2-9)

= (x-1)(x+2)(x²-32)

= (x-1)(x+2)(x+3)(x-3)

Now P(x)

(x-1)(x+2)(x+3)(x-3)=0

= (x-1)(x+2)(x+3)(x-3)

⇒ x-1=0 x+2=0

⇒ x+3=0 X-3=0

⇒ x=1 x=-2 x=-3 X=3

Hence, other Zeroes are 3 and -3.

Question 16. Find all Zeroes of x + x3-23x=-3x+60, if it is given that two of its Zeroes are √3 and -√3.

Solution:

Let P(x) = x2+x3-28x=-3x+60

√3 and -√3 are Zeroes of P(x).

(x−√3)(x+√3) = x2= 3 is a factor of P(x).

P(x) = x²+x³-x²=3x+60 = (x²=3)(x²+11-20)

= (x²-3) [x²+5x-4x-20]

=(x²-3)(x(x+5)-4(x+5))

= (x²-3)(x-4)(x+5)

The other Zeroes are given by

x-4=0 or x+5=0

⇒ x=4 Οr x=-5

Hence, other Zeroes are 4 and -5.

Question 17. Find Zeroes of the polynomial f(x) = x² – 13x² + 32x – 60, if it is given that the Product of its two Zeroes is 10.

Solution: Let α, B, be Zeroes of the given polynomial p(x), Such that &p=10-)(1)

⇒ \(\alpha+\beta+\gamma=\frac{-(-13)}{1}=13 \longrightarrow(2)\)

⇒ \(\alpha \beta+\beta 1+\alpha \gamma=\frac{32}{1}=32 \longrightarrow(3)\)

⇒ \(\alpha \beta \gamma=-\frac{(-60)}{1}=60 \longrightarrow(4)\)

From (1) and (4)

10s=60

⇒ \(\gamma=\frac{60}{10} \Rightarrow \gamma=6 \rightarrow(5)\)

Put 7=6 in (2), we get α+3 +6 = 13

α+B=7

Now, (α-B)² = (x+3)=4xß

= (7)=4(10)

= 49-40

= 9

α-B = ± 3 —– (6)

Solving (5) and (6), we get

α=2,B=5 or α=2, B=5 and 1=6.

So, Zeroes are 2,5 and 6.

Question 18. What must be added to P(x) = 4x² – 5x = 39x = 46x = -2, so that the resulting Polynomial is divisible by g(x) = 4x² + 7x + 2 ?

Solution:

P(x) = 4×4 -5x³-39x²– 46x-2

9(x)= 4x²+7x+2