Question

y = kx2+(3K+2)x+R
Q. Find the value of y for which
Gjyso, XER
yo & XER.
(1) ys o for atleast one x.
ON) yco
1)​

Answer 1

Answer:

2x

2

+kx+3=0 ---(1)

if eq (1)has equal roots then

Discriminant (1)=

b

2

−4ac

=0

b

2

=4ac

k

2

=4×2×3

k

2

=24

k=

24

=±2

6

∴ for k = ±2

16

given equation has equal roots.

Step-by-step explanation:

PLEASE MARK AS BRAINLIST HOPE IT IS HELPFUL FOR YOU