Ex-8.3 Quadratic Equations (Part - 1), 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.