Question

Solve the given quadratic equation:
2x^3 - 11x^2 +44x +27 , if x = 25/3-4i

Answer 1

Answer:

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R.D. Sharma Solutions»R.D. Sharma Class 10 Solutions»Chapter 8 Quadratic Equations»Quadratic Equations Exercise 8.3

Chapter 8: Quadratic Equations Exercise – 8.3

Question: 1

Find the roots of the equation (x - 4) (x + 2) = 0

Solution:

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

Solution:

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

Solution:

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.

Solution:

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

Solution:

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