Question

for what value of k does (k-12)x²+2(k-12)x+2=0 have equal roots?

Answer 1

#AlexaRousey here!!

= (k-12)x²+2(k-12)x+2=0

To find : value of k such that it have equal roots.

A polynomial have equal roots when D= 0.

Here a = k-12,
b = 2(k-12) nd c = 2.

=) D = 0

=) b^2 - 4ac = 0

=) {2(k-12)}^2 - 4(k-12)(2) = 0

=) (2k-24)^2 - 4(2k-24) = 0

=) (2k - 24) (2k-24 - 4) = 0

=) (2k-24) (2k-28) = 0

=) 2k = 24 or 2k = 28

=) k = 12 or k= 14

Here k can't equal to 12 because of inequality.

Hence k = 14.

Thanks!!