Question

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

Answer 1

Step 1: Determining the discriminant $\left(D\right)$ of the quadratic equation

Given that, equation $\left(k+4\right)x^2+\left(k+1\right)x+1=0$ has equal roots.

The discriminant of the quadratic equation is determined by using the formula:

$D=B^2-4AC$

For equation $\left(k+4\right)x^2+\left(k+1\right)x+1=0$ the value $A=\left(k+4\right),B=\left(k+1\right)$ and $C=1$.

Substituting the values of $A$, $B$ and $C$ in discriminant formula.

$\begin{array}{rcl}D& =& B^2-4AC\\ & =& {\left(k+1\right)}^2-4\left(k+4\right)\left(1\right)\\ & =& k^2+2k+1-4k-16\\ & =& k^2-2k-15\end{array}$

Step 2: Determining the values of $k$

According to the nature of roots of a quadratic equation if the discriminant $D=0$, then the quadratic equation has real and equal roots.

By nature of roots of equation,

$\begin{array}{rcl}D& =& 0\\ & \Rightarrow & k^2-2k-15=0\end{array}$

Using factorization method

$\begin{array}{rcl}{k}^2-2k-15& =& 0\\ & \Rightarrow & k^2-5k+3k-15=0\\ & \Rightarrow & k\left(k-5\right)+3\left(k-5\right)=0\\ & \Rightarrow & k=-3,5\end{array}$

Hence, for $k=-3,5$ the equation $\left(k+4\right)x^2+\left(k+1\right)x+1=0$ has equal roots.