Question

For what value of $k$, are the roots of the quadratic equation $(k+4)x^2+(k+1)x+1=0$ equal?

Answer 1

Roots are equal$\Rightarrow$ Discriminant$=0$

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

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

$\Rightarrow k^2+2k+1-4k-16=0$

$\Rightarrow k^2-2k+1-16=0$

$\Rightarrow {(k-1)}^2-4^2=0$

$\Rightarrow (k-1-4)(k-1+4)=0$

$\Rightarrow (k-5)(k+3)=0$

$\therefore k=5,-3$