Question

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

Answer 1

Discriminant,

$D=b^2-4ac$

$D=(k+1{)}^2-4(k+4)$

$D=k^2-2k-15$

$D=(k-5)(k+3)$

If the roots of the given equation are real and equal, then,

$D=0$

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

$k=5,-3$