Question

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

Answer 1

In equation $a{x}^{2}bx+c=0$

If roots are equal, then

$b^2-4ac=0$

In given equation,

$(k+4)x^2+(k+1)x+1=0$

Where $a=(k+4),b=(k+1)andc=1$

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

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

$\Rightarrow k^2-2k-15=0$

$\Rightarrow k^2-5k+3k-15=0$

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

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

If $k-5=0$ then $k=5$

If $k+3=0$ then $k=-3$