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.

HOPE THIS ANSWER WILL HELP YOU... ..

Answer 2

Step-by-step explanation:

kx^2+6x+4k=0

here,a=k

b=6

c=4k

we know.sum of roots=  -b/a

so, here sum of the roots= -6/k

& product of roots=c/a

here, product of roots=4k/k=4

A.C.Q, -6/K=4

4k=-6

k= -3/2