Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 PDF Download

Question 1: Find the roots of the equation (x – 4) (x + 2) = 0

Sol:

The given equation is (x – 4) (x + 2) = 0

Either x – 4 = 0 therefore x = 4

Or, x + 2 = 0 therefore x = - 2

The roots of the above mentioned quadratic equation are  4 and - 2 respectively.

Question 2: Find the roots of the equation (2x + 3) (3x – 7) = 0

Sol:

The given equation is (2x + 3) (3x – 7) = 0.

Either 2x + 3 = 0, therefore x = −3/2

Or, 3x - 7 = 0, therefore x = 7/3

The roots of the above mentioned quadratic equation are x = −3/2 and x = 7/3 respectively.

Question 3: Find the roots of the quadratic equation 3x² - 14x - 5 = 0

Sol:

The given equation is 3x² - 14x - 5 = 0

= 3x² - 14x - 5 = 0

= 3x² - 15x + x - 5 = 0

= 3x(x - 5) + 1(x - 5) = 0

= (3x + 1)(x - 5) = 0

Either 3x + 1 = 0 therefore x = −1/3

Or, x - 5 = 0 therefore x = 5

The roots of the given quadratic equation are 5 and x = −1/3 respectively.

Question 4: Find the roots of the equation 9x² - 3x - 2 = 0.

Sol:

The given equation is 9x² - 3x - 2 = 0.

= 9x² - 3x - 2 = 0.

= 9x²  - 6x + 3x - 2 = 0

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

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

Either, 3x - 2 = 0 therefore x = 2/3

Or, 3x + 1 = 0 therefore x = −1/3

The roots of the above mentioned quadratic equation are  x = 2/3 and x = −1/3 respectively.

Question 5: Find the roots of the quadratic equation 

Sol:

The given equation is   

Cancelling out the like terms on both the sides of the numerator. We get,

= x² + 4x - 5 = 7

= x² + 4x - 12 = 0

= x² + 6x - 2x - 12 = 0

= x(x + 6) - 2(x - 6) = 0

= (x + 6)(x - 2) = 0

Either x + 6 = 0

Therefore x = - 6

Or, x - 2 = 0

Therefore x = 2

The roots of the above mentioned quadratic equation are 2 and - 6 respectively.

Question 6: Find the roots of the equation 6x² + 11x + 3 = 0.

Sol:

The given equation is 6x² + 11x + 3 = 0.

= 6x² + 11x + 3 = 0.

= 6x²  + 9x + 2x + 3 = 0

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

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

Either, 2x + 3 = 0 therefore x = −3/2

Or, 3x + 1 = 0 therefore x = −1/3

The roots of the above mentioned quadratic equation are x = −3/2 and  x = −1/3 respectively .

Question 7: Find the roots of the equation 5x² - 3x - 2 = 0

Sol:

The given equation is 5x² - 3x - 2 = 0.

= 5x² - 3x - 2 = 0.

= 5x²  - 5x + 2x - 2 = 0

= 5x(x - 1) + 2(x - 1) = 0

= (5x + 2) (x - 1) = 0

Either 5x + 2 = 0 therefore x = −2/5

Or, x - 1 = 0 therefore x = 1

The roots of the above mentioned quadratic equation are 1 and x = −2/5 respectively.

Question 8: Find the roots of the equation 48x² - 13x - 1 = 0

Sol:

The given equation is 48x² - 13x - 1 = 0.

= 48x² - 13x - 1 = 0.

= 48x² - 16x + 3x - 1 = 0.

= 16x (3x - 1) + 1(3x - 1) = 0

= (16x + 1)(3x - 1) = 0

Either 16x + 1 = 0 therefore x = −1/16

Or, 3x - 1 = 0 therefore x = 1/3

The roots of the above mentioned quadratic equation are  x = −1/16

And x = 1/3  respectively.

Question 9: Find the roots of the equation 3x² = - 11x - 10

Sol:

The given equation is 3x² = - 11x - 10

= 3x² = - 11x - 10

= 3x² + 11x + 10 = 0

= 3x² + 6x + 5x + 10 = 0

= 3x(x + 2) + 5(x + 2) = 0

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

Either 3x + 2 = 0 therefore x = −2/3

Or, x + 2 = 0 therefore x = - 2

The roots of the above mentioned quadratic equation are x = −2/3 and - 2 respectively.

Question 10: Find the roots of the equation 25x(x + 1) = - 4

Sol:

The given equation is 25x(x + 1) = 4

= 25x(x + 1) = - 4

= 25x² + 25x + 4 = 0

= 25x² + 20x + 5x + 4 = 0

= 5x (5x + 4) + 1(5x + 4) = 0

= (5x + 4)(5x + 1) = 0

Either 5x + 4 = 0 therefore x = −4/5

Or, 5x + 1 = 0 therefore x = −1/5

The roots of the quadratic equation are  x = −4/5 and  x = −1/5 respectively.

Question 12: Find the roots of the quadratic equation 1/x− = 3

Sol:

The given equation is 1/x− = 3

=  1/x− = 3

Cross multiplying both the sides. We get,

= 2 = 3x(x - 2)

= 2 = 3x² - 6x

= 3x² - 6x - 2 = 0

= 3x² - 3x - 3x - 2 = 0

= 3x²−(3 + √3)x−(3−√3)x + [(√32)−12]

=  3x²−(3 + √3)x−(3−√3)x + [(√3²)−1²][(√3²)−1²]

=  √3²x²−√3(√3 + 1)x−√3(√3−1)x + (√3 + 1)(√3−1)

=  √3x(√3 + 1)x−(√3x−(√3 + 1))(√3−1)

=  (√3x−√3−1)(√3x−√3 + 1)(√3−1)

Either =

Therefore x =

Or, (√3x−√3 + 1)(√3−1)

Therefore, x =

The roots of the above mentioned quadratic equation are x =  and x =  respectively.

Question 13: Find the roots of the quadratic equation x−1/x = 3

Sol:

The given equation is x−1/x = 3

=  x−1/x = 3

= x² - 1 = 3x

= x² - 1 - 3x = 0

Either 

Therefore 

Therefore 

The roots of the above mentioned quadratic equation are  and  respectively.

Question 14: Find the roots of the quadratic equation 

Sol:

The given equation is 

Cancelling out the like numbers on both the sides of the equation

= x² - 3x - 28 = - 30

= x² - 3x - 2 = 0

= x² - 2x - x - 2 = 0

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

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

Either x - 2 = 0

Therefore x = 2

Or, x - 1 = 0

Therefore x = 1

The roots of the above mentioned quadratic equation are 1 and 2 respectively.

Question 15: Find the roots of the quadratic equation a²x² - 3abx + 2b² = 0

Sol:

The given equation is a²x² - 3abx + 2b² = 0

= a²x² - 3abx + 2b² = 0

= a²x² - abx - 2abx + 2b² = 0

= ax(ax - b) - 2b(ax - b) = 0

= (ax - b)(ax - 2b) = 0

Either ax - b = 0 therefore x = b/a

Or, ax - 2b = 0 therefore x = 2b/a

The roots of the quadratic equation are x = 2b/a and x = b/a respectively.

Question 16: Find the roots of the 4x² + 4bx - (a² - b²) = 0

Sol:

- 4(a² - b²) = - 4(a - b)(a + b)

= - 2(a - b) * 2(a + b)

= 2(b - a) * 2(b + a)

= 4x² + (2(b - a) + 2(b + a)) – (a - b)(a + b) = 0

=    4x²   + 2(b - a)x + + 2(b + a)x + (b - a)(a + b) = 0

=   2x(2x + (b - a)) + (a + b)(2x + (b - a)) = 0

= (2x + (b - a))(2x + b + a) = 0

Either, (2x + (b - a)) = 0

Therefore x =

Or, (2x + b + a) = 0

Therefore x =

The roots of the above mentioned quadratic equation are x = and x =  respectively.

Question 17: Find the roots of the equation ax² + (4a² - 3b)x - 12ab = 0

Sol:

The given equation is ax² + (4a² - 3b)x - 12ab = 0

= ax² + (4a² - 3b)x - 12ab = 0

= ax² + 4a²x - 3bx - 12ab = 0

= ax(x - 4a) – 3b(x - 4a) = 0

= (x - 4a)(ax - 4b) = 0

Either x - 4a = 0

Therefore x = 4a

Or, ax - 4b = 0

Therefore x = 4ba

The roots of the above mentioned quadratic equation are x = 4b/a and 4a respectively.

Question 18: Find the roots of 

Sol:

The given equation is 

= = (x + 3)(2x - 3) = (x + 2){3x - 7)

= 2x² - 3x + 6x - 9 = 3x² - x - 14

= 2x² + 3x - 9 = 3x² - x - 14

= x² - 3x - x - 14 + 9 = 0

= x² - 5x + x - 5 = 0

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

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

Either x - 5 - 0 or x + 1 = 0

x = 5 and x = - 1

The roots of the above mentioned quadratic equation are 5 and - 1 respectively.

Question 19: Find the roots of the equation 

Sol:

The given equation is 

= 3(4x² - 19x + 20) = 25(x² - 7x + 12)

= 12x² - 57x + 60 = 25x² – 175x + 300

= 13x² - 78x - 40x + 240 = 0

= 13x² - 118x + 240 = 0

= 13x² - 78x - 40x + 240 = 0

= 13x(x - 6) - 40(x - 6) = 0

= (x - 6)(13x - 40) = 0

Either x - 6 = 0 therefore x = 6

Or , 13x - 40 = 0 therefore x =  40/13

The roots of the above mentioned quadratic equation are 6 and 40/13   respectively.

Question 20: Find the roots of the quadratic equation 

Sol:

The given equation is 

= 4(2x² + 2) = 17(x² - 2x)

= 8x² + 8 = 17x² - 34x

= 9x² - 34x - 8 = 0

= 9x² - 36x + 2x - 8 = 0

= 9x(x - 4) + 2(x - 4) = 0

= (9x + 2)(x - 4) = 0

Either 9x + 2 = 0 therefore x = −2/9

Or, x - 4 = 0 therefore x = 4

The roots of the above mentioned quadratic equation are x = −2/9 and 4 respectively.

Question 21: Find the roots of the quadratic equation 

Sol:

The equation is 

= x(3x - 5) = 6(x² - 3x + 2)

= 3x² - 5x = 6x² - 18x + 12

= 3x² - 13x + 12 = 0

= 3x² - 9x - 4x + 12 = 0

= 3x(x - 3) - 4(x - 3) = 0

= (x - 3)(3x - 4) = 0

Either x - 3 = 0 therefore x = 3

Or, 3x - 4 = 0 therefore 4/3

The roots of the above mentioned quadratic equation are 3 and 4/3 respectively.

Question 22: Find the roots of the quadratic equation 

Sol:

The equation is 

= 6(4x) = 5(x² - 1)

= 24x = 5x² - 5

= 5x² - 24x - 5 = 0

= 5x² - 25x + x - 5 = 0

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

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

Either x - 5 = 0

Therefore x = 5

Or, 5x + 1 = 0

Therefore x = −1/5

The roots of the above mentioned quadratic equation are x = −1/5 and 5 respectively.

Question 23: Find the roots of the quadratic equation 

Sol:

The equation is 

= 2(5x² + 2x + 2) = 5(2x² - x - 1)

= 10x² + 4x + 4 = 10x² - 5x - 5

Cancelling out the equal terms on both sides of the equation. We get,

= 4x + 5x + 4 + 5 = 0

= 9x + 9 = 0

= 9x = - 9

X = - 1

X = - 1 is the only root of the given equation.

Question 24: Find the roots of the quadratic equation = 1−2x

The given equation is  = 1−2x

=  = 1−2x

  = 1−2x

= m²x² + 2mnx + (n² - mn) = 0

Now we solve the above quadratic equation using factorization method

Therefore

Now, one of the products must be equal to zero for the whole product to be zero for the whole product to be zero. Hence, we equate both the products to zero in order to find the value of x .

Therefore,

Or

The roots of the above mentioned quadratic equation are respectively.

Question 25: Find the roots of the quadratic equation 

Sol:

The given equation is 

= (2x² - 2x(a + b) + a² + b²)ab = (a² + b²)(x² - (a + b)x + ab)

= (2abx² - 2abx(a + b) + ab(a² + b²)) = (a² + b²)(x² - (a² + b²)(a + b)x + (a² + b²)(ab)

= (a² + b² - 2ab)x - (a + b)(a² + b² - 2ab)x = 0

= (a - b)²x² - (a + b)(a + b)²x² = 0

= x(a - b)²(x - (a + b)) = 0

= x(x - (a + b)) = 0

Either x = 0

Or, (x - (a + b)) = 0

Therefore x = a + b

The roots of the above mentioned quadratic equation are 0 and a + b respectively.

Question 26: Find the roots of the quadratic equation 

Sol:

The given equation is 

Cancelling out the like terms on both the sides of numerator and denominator. We get,

= (x - 1)(x - 4) = 18

= x² - 4x - x + 4 = 18

= x² - 5x - 14 = 0

= x² - 7x + 2x - 14 = 0

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

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

Either x - 7 = 0

Therefore x = 7

Or, x + 2 = 0

Therefore x = - 2

The roots of the above mentioned quadratic equation are 7 and - 2 respectively.

Question 27: Find the roots of the quadratic equation 

Sol:

The given equation is 

= (x - c)(ax - 2ab + bx) = 2c(x² - bx - ax + ab)

= (a + b)x² - 2abx - (a + b)cx + 2abc = 2cx² - 2c(a + b)x + 2abc

Question 28: Find the roots of the Question x² + 2ab = (2a + b)x

Sol:

The given equation is x² + 2ab = (2a + b)x

= x² + 2ab = (2a + b)x

= x² - (2a + b)x + 2ab = 0

= x² - 2ax - bx + 2ab = 0

= x(x - 2a) - b(x - 2a) = 0

= (x - 2a)(x - b) = 0

Either x - 2a = 0

Therefore x = 2a

Or, x - b = 0

Therefore x = b

The roots of the above mentioned quadratic equation are 2a and b respectively.

Question 29: Find the roots of the quadratic equation (a + b)²x² - 4abx - (a - b)²  = 0

Sol:

The given equation is (a + b)²x² - 4abx - (a - b)²  = 0

= (a + b)²x² - 4abx - (a - b)²  = 0

= (a + b)²x² - ((a + b)² –(a - b)² )x - (a - b)²  = 0

= (a + b)²x² - (a + b)² x + (a - b)² x - (a - b)²  = 0

= (a + b)²x(x - 1) (a + b)² (x - 1) = 0

= (x - 1) (a + b)²x + (a + b)²) = 0

Either x - 1 = 0

Therefore x = 1

Or, (a + b)²x + (a + b)²) = 0

Therefore 

The roots of the above mentioned quadratic equation are and 1 respectively .

Question 30: Find the roots of the quadratic equation a(x² + 1) - x(a² + 1) = 0

Sol:

The given equation is a(x² + 1) - x(a² + 1) = 0

= a(x² + 1) - x(a² + 1) = 0

= ax² + a - a²x - x = 0

= ax(x - a) - 1(x - a) = 0

= (x - a)(ax - 1) = 0

Either x - a = 0

Therefore x = a

Or, ax - 1 = 0

Therefore x = 1/a

The roots of the above mentioned quadratic equation are  ( a) and  x = 1/a respectively.

Question 31: Find the roots of the quadratic equation x² + (a + 1/a)x + 1 = 0.

Sol:

The given equation is x² + (a + 1/a)x + 1 = 0

=  x² + (a + 1/a)x + 1 = 0

=  x² + ax + x/a + (a×1/a) = 0

=  x(x + a) + 1/a(x + a) = 0

=  (x + a)(x + 1/a) = 0

Either x + a = 0

Therefore x = - a

Or , (x + 1/a) = 0

Therefore x = 1/a

The roots of the above mentioned quadratic equation are x = 1/a and –a respectively.

Question 32: Find the roots of the quadratic equation abx² + (b² - ac)x - bc = 0

Sol:

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

= abx² + (b² - ac)x - bc = 0

= abx² + b²x - acx - bc = 0

= bx (ax + b) - c (ax + b) = 0

= (ax + b)(bx - c) = 0

Either, ax + b = 0

Therefore x = −b/a

Or, bx - c = 0

Therefore x = c/b

The roots of the above mentioned quadratic equation are x = c/b and x = −b/a respectively.

Question 33: Find the roots of the quadratic equation a²b²x² + b²x - a²x - 1 = 0

Sol:

The given equation is a²b²x² + b²x - a²x - 1 = 0

= a²b²x² + b²x - a²x - 1 = 0

= b²x(a²x + 1) - 1(a²x + 1)

= (a²x + 1)( b²x - 1) = 0

Either (a²x + 1) = 0

Therefore x = 

Or, ( b²x - 1) = 0

Therefore x =

The roots of the above mentioned quadratic equation are x =  and x =  respectively.