Question

For what value of k the equation kx(x-2) +6=0 has two real equal roots?

Answer 1

SOLUTION

kx(x-2)+6=0

=) kx^2 -2kx+6= 0

Comparing it with Ax^2 Bx +C= 0

A= k, B = -2k and C= 6

=) D= B^2- 4AC

=) 4k^2 -24k

=) 4k(k-6)

D=0

=)4k(k-6)=0

=) k=0 or k= 6

HOPE IT HELPS ☺️

Answer 2

Answer:

3/2

Step-by-step explanation:

for real and equal root

D=0

kx^2-2kx+6=0

(4k)^2 - 4(k)(6)=0

16(k)^2 - 24k=0

2(k)^2 - 3k=0

(2k-3)=0

k=3/2