Question

The quadratic equation (3k-2)x^2 +12x+3(k+1) =0 has equal roots. Find the two possible values of k.

Answer 1

$x_1=x_2 \Rightarrow \Delta=0\\ \Delta=12^2-4\cdot(3k-2)\cdot3(k+1)\\ \Delta=144-(36k-24)(k+1)\\ \Delta=144-36k^2-36k+24k+24\\ \Delta=-36k^2-12k+168\\ -36k^2-12k+168=0\\ -3k^2-k+14=0\\ -3k^2+6k-7k+14=0\\ -3k(k-2)-7(k-2)=0\\ -(3k+7)(k-2)=0\\ k=-\frac{7}{3} \vee k=2$