Math, asked by jiocompany, 1 year ago

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}

Answers

Answered by mathsdude85
0

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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