In this page nature of roots solution2 we are going to see solutions of some practice questions of the worksheet nature of roots.

Determine the nature of the roots of the equation.

(v) ( 3/5) x² - (2/3) x + 1 = 0

To get the values of a , b and c we have to compare the given equation with the general form of quadratic equation a x² + b x + c = 0

a = 3/5 b = -2/3 and c = 1

Discriminant ∆ = b² - 4 ac

= (-2/3)² - 4 (3/5) (1)

= 4/9 - 12/5

= (20-108)/45

= -88/45

The value of ∆ is -88/45 that is ∆ < 0 .Hence the roots are imaginary (not real).

(vi) (x - 2a) (x - 2b) = 4 a b

Now we have to make the given equation in the form of a x² + b x + c = 0

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

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

x² + x (2 b - 2 a) = 0

from this we come to know the values of a ,b and c

a = 1 b = 2 b - 2a c = 0

Discriminant ∆ = b² - 4 ac

= (2b - 2a)² - 4 (1) (0)

= [2 ( b -a)]² - 0

= [2 (b - a)]²

The value of ∆ is [2 (b - a)]² that is ∆ > 0 and it is a perfect suqare. Hence the roots are real,unequal and irrational.

(2) Find the values of k for which the roots are real and equal of each of the following equations.

(i) 2 x² - 10 x + k = 0

Since the roots are real and equal the value of ∆ = 0

a = 2 b = -10 c = k

∆ = b² - 4 a c

(-10)² - 4 (2) (k) = 0

100 - 8 k = 0

- 8 k = -100

8 k = 100

k = 100/8

k = 25/2

nature of roots solution2 nature of roots solution2

- Back to worksheet
- Practical problems in quadratic equation
- Framing quadratic equation from roots
- Square root
- Solving linear equation in cross multiple method
- Solving linear equations in elimination method