Question

4.
Find the value of k, if the roots of the following equations are equal.
(2k-1) x2 + 3kx + 3 = 0
(ii) x2 + 2(k + 2) x + (3k + 4) = 0
(iii) (3k + 2) x2 - 5(ky+1)x+(2k+ 3) = 0​

Answer 1

(k-4)x²+2(k-4)x+2=0

IF BOTH THE ROOTS ARE EQUAL;THEN DISCRIMINANT ⇒b²=4ac

then a=k-4; b=2(k-4) ;c=2; substituting the values

(2k-8)²=4×(k-4)×2

4k²+64-32k=8k-32

4k²-40k+96=0

k²-10k+24=0

(k-4)(k-6)=0

k=4,6