Question

For what values of k does (k-12)x^2+2(k-12)x+2=0 has equal roots?

Answer 1

[tex]( 2(k-12))^{2}-4(k-12)(2)=0 [/tex] ⇒condition for equal roots
⇒(k-12)²-(2k-24)=0
⇒k²+144-24k-2k+24=0
⇒k²-26k+168=0
Now, roots of this equation in k are:-
$\alpha = \frac{-b+ \sqrt{b^{2} -4ac} }{2a} , \beta =\frac{-b- \sqrt{b^{2} -4ac} }{2a}$
⇒$\alpha = \frac{26+2}{2}$
⇒$\alpha =14$
now,
⇒$\beta = \frac{26-2}{2}$
⇒$\beta =12$

Therefore k can be either 14 or 12