CBSE Solutions For Class 10 Mathematics Chapter 4 Quadratic Equations

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 \(\frac{1}{3}\)