Question

Find the value of k for which the equadratic equation (k-2)x square+2(2k-3)x+(5k-6)=0 as real and equal roots

Answer 1

Answer:

Step-by-step explanation:

K-2x^2+2(2K-3)+5K-6=0

Equal and real roots

b^2-4ac

=2(2K-3)^2 -- 4[(K-2)(5K-6)]

=(4K-6)^2--(4K-8)(5K-6)

=16K^2+36-48K--20K^2+40K+24K--48

=-4K^2-16K-8

=K^2-4K-2

=K^2-2K-2K-2