Question

Find the values of k so that the equation $x^2-2x(1+3k)+7(3+2k)=0$ has real and equal roots.

Answer 1

The given equation is $x^2-2x(1+3k)+7(3+2k)=0$.

It has real and equal roots. So, $b^2-4ac=0$.

$[-2(1+3k){]}^2-4(1\left)\right(7\left)\right(3+2k)=0$

$4(1+9k^2+6k)-4(21+14k)=0$

$4+36k^2+24k-84-56k=0$

$36k^2-32k-80=0$

$9k^2-8k-20=0$

$9k^2-18k+10k-20=0$

$9k(k-2)+10(k-2)=0$

$(k-2)(9k+10)=0$

$k=2,\frac{-10}{9}$