CBSE Solutions For Class 10 Mathematics Chapter 4 Quadratic Equations

CBSE Solutions For Class 10 Mathematics Chapter 4 Quadratic Equations

Question 1. Which of the following are quadratic equations?

1.  X28x+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.

CBSE Class 10 Maths Chapter 4 Quadratic Equation

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.

CBSE Class 10 Maths Chapter 4 Quadratic Equation

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.

CBSE Solutions For Class 10 Mathematics Chapter 4 Quadratic Equations

CBSE Class 10 Maths Chapter 4 Quadratic Equation

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.

CBSE Class 10 Maths Chapter 4 Quadratic Equation

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.

Class 10 Quadratic Equations NCERT 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.

Class 10 Quadratic Equations NCERT 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.

Class 10 Quadratic Equations NCERT 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 \(x=\frac{-3}{4}\)

Quadratic Equations Class 10 Exercise Solutions

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:

Quadratic Equations Class 10 Exercise 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.

Quadratic Equations Class 10 Exercise 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.

Quadratic Equations Class 10 Exercise Solutions

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

Quadratic Equations Class 10 Exercise Solutions

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

Solution:

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

4x2-3x+1

4x2-37-1=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.

CBSE Class 10 Quadratic Equations Important Questions

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.

CBSE Class 10 Quadratic Equations Important Questions

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.

CBSE Class 10 Quadratic Equations Important Questions

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.

CBSE Class 10 Quadratic Equations Important Questions

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.

CBSE Class 10 Quadratic Equations Important Questions

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.

CBSE Class 10 Quadratic Equations Important Questions

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.

CBSE Class 10 Quadratic Equations Important Questions

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.

CBSE Class 10 Quadratic Equations Important Questions

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.

NCERT Solutions for Class 10 Maths Chapter 4

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.

NCERT Solutions for Class 10 Maths Chapter 4

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.

NCERT Solutions for Class 10 Maths Chapter 4

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.

NCERT Solutions for Class 10 Maths Chapter 4

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.

NCERT Solutions for Class 10 Maths Chapter 4

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.

Nature of Roots of Quadratic Equations Class 10

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.

Nature of Roots of Quadratic Equations Class 10

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.

Nature of Roots of Quadratic Equations Class 10

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

Nature of Roots of Quadratic Equations Class 10

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.

Nature of Roots of Quadratic Equations Class 10

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.

Nature of Roots of Quadratic Equations Class 10

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.

Nature of Roots of Quadratic Equations Class 10

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

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