Question

Find the values of $k$ for the following quadratic equation, so that they have two real and equal roots:$(k+4)x^2+(k+1)x+1=0$

• A. $k=5,-3$
• B. $k=3,5$
• C. $k=3,-5$
• D. $k=-3,-5$

Answer 1

The correct option is A $k=5,-3$

Given quadratic equation is $(k+4)x^2+(k+1)x+1=0$ and roots are equal and zero.

For a quadratic equation $a{x}^2+bx+c=0$ roots are real and equal, if $D=b^2-4ac=0$.

Here $a=(k+4),b=(k+1),c=1$

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

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

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

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

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

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

$\Rightarrow k-5=0$ and $k+3=0$

$\Rightarrow k=5$ and $k=-3$