Question

Find the value of $a$ for which the equation $(k-1)x^2+(k+4)x+k+7=0$ has equal roots.

Answer 1

For equal roots it must have $D=0$ where $D=b^2-4ac$

$\Rightarrow b^2-4ac=0$

$\Rightarrow$ $(k+4{)}^2-4\times (k-1\left)\right(k+7)=0$

$\Rightarrow$ $(k^2+16+8k)-4(k^2+7k-k-7)=0$

$\Rightarrow$ $k^2+16+8k-4k^2-28k+4k+28=0$

$\Rightarrow$ $-3k^2-16k+44=0$

$\Rightarrow$ $3k^2+16-44=0$

By solving the equation we get the value $k=2,-\mathrm{7.333.....}$