Class 10 Exam  >  Class 10 Notes  >  Extra Documents, Videos & Tests for Class 10  >  RD Sharma Solutions - Ex-8.13 Quadratic Equations, Class 10, Maths

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

Sol:

The given equation is Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10  

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

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

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

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

= 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.

= 6x + 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−Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 = 3

Sol:

The given equation is 1/x−Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 = 3

=  1/x−Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 = 3

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

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

Therefore x = Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10

Or, (√3x−√3 + 1)(√3−1)

Therefore, x = Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10

The roots of the above mentioned quadratic equation are x = Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 and x = Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 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

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

= x2 - 1 = 3x

= x2 - 1 - 3x = 0

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

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

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

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

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

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

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

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

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

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

The roots of the above mentioned quadratic equation are Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 and Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 respectively.

 

Question 14: Find the roots of the quadratic equation Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10

Sol:

The given equation is Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10

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

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

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

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

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

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

Therefore x =Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10

The roots of the above mentioned quadratic equation are x = Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10and x = Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 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 Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10

Sol:

The given equation is Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10

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

Sol:

The given equation is Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10

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

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

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

Sol:

The given equation is Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10

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

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

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

Sol:

The equation is Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10

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

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

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

Sol:

The equation is Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10

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

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

Sol:

The equation is Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10

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

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

= 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 Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10= 1−2x

The given equation is Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 = 1−2x

Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 = 1−2x

 Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 = 1−2x

= m2x2 + 2mnx + (n2 - mn) = 0

Now we solve the above quadratic equation using factorization method

Therefore

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

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

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

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

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,

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

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

Or

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

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

The roots of the above mentioned quadratic equation are Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10respectively.

 

Question 25: Find the roots of the quadratic equation Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10

Sol:

The given equation is Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10

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

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

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

Sol:

The given equation is Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10

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

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

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

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

 

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

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

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

Sol:

The given equation is Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10

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

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

= (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) = 0

Sol:

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

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

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

= (a + b)2x2 - (a + b)x + (a - b)x - (a - b) = 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 Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10

The roots of the above mentioned quadratic equation are Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10and 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 x+ (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 = Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10

Or, ( b2x - 1) = 0

Therefore x = Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10

The roots of the above mentioned quadratic equation are x = Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 and x = Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 respectively.

The document Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 is a part of the Class 10 Course Extra Documents, Videos & Tests for Class 10.
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FAQs on Ex-8.13 Quadratic Equations, Class 10, Maths RD Sharma Solutions - Extra Documents, Videos & Tests for Class 10

1. What are the different methods to solve quadratic equations?
Ans. There are mainly three methods to solve quadratic equations: factorization method, completing the square method, and using the quadratic formula.
2. How can I determine the nature of roots of a quadratic equation?
Ans. The nature of roots of a quadratic equation can be determined by calculating the discriminant. If the discriminant is greater than zero, then the equation has two distinct real roots. If the discriminant is equal to zero, then the equation has two identical real roots. And if the discriminant is less than zero, then the equation has no real roots.
3. Can a quadratic equation have more than two solutions?
Ans. No, a quadratic equation can have a maximum of two solutions. This is because a quadratic equation represents a parabola, which intersects the x-axis at most two times. Therefore, it can have either two real solutions or no real solutions.
4. How can I find the roots of a quadratic equation if it cannot be factorized?
Ans. If a quadratic equation cannot be factorized, we can still find its roots using the quadratic formula. The quadratic formula states that for any quadratic equation of the form ax^2 + bx + c = 0, the roots can be calculated as x = (-b ± √(b^2 - 4ac)) / 2a.
5. Can a quadratic equation have complex roots?
Ans. Yes, a quadratic equation can have complex roots. If the discriminant of the quadratic equation is less than zero, then the roots are complex conjugates. Complex roots are in the form of a + bi, where 'a' and 'b' are real numbers and 'i' is the imaginary unit (defined as √(-1)).
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