Question

If the sum and product of the roots of the equation kx² + 6x + 4k = 0 are real, then k =
(a) $\frac{-3} {2}$
(b)$\frac{3} {2}$
(c) $\frac{2} {3}$
(d)$\frac{-2} {3}$

Answer 1

SOLUTION :  

Option (a) is correct : - 3/2

Given : kx² + 6x + 4k = 0 and sum and product of roots are equal.

On comparing the given equation with ax² + bx + c = 0  

Here, a = k , b = 6 , c =  4k

Sum of roots = - b/a

Sum of roots = - 6/k

Product of roots = c/a

Product of roots = 4k/k

Product of roots = 4

Given : sum of roots = Product of roots

-6/k = 4

4k = - 6

k = -6/4

k = - 3/2

Hence, the value of k is - 3/2.

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Answer 2

Given : kx² + 6x + 4k = 0 and sum and product of roots are equal.

On comparing the given equation with ax² + bx + c = 0  

Here, a = k , b = 6 , c =  4k

Sum of roots = - b/a

Sum of roots = - 6/k

Product of roots = c/a

Product of roots = 4k/k

Product of roots = 4

Given : sum of roots = Product of roots

-6/k = 4

4k = - 6

k = -6/4

k = - 3/2

Hence, the value of k is - 3/2.