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

Step 1: Using the discriminant, compare the general quadratic equation

Discriminant $D=B^2-4AC$ compared with the equation $A{x}^2+Bx+C=0$.

So, $A=k+4,B=k+1$ and, $C=1$

For roots to be equal, $D=0$.

Equate the values of $A,B$ and $C$ in the equation $D=B^2-4AC$

$$\begin{gathered} \Rightarrow {(k+1)}^2-4(k+4)\left(1\right)=0 \\ \Rightarrow k^2+2k+1-4k-16=0 \\ \Rightarrow k^2-2k-15=0 \end{gathered}$$

Step 2: Find the value of $k$:

After simplifying,

$$\begin{gathered} \Rightarrow k^2-5k+3k-15=0 \\ \Rightarrow k(k-5)+3(k-5)=0 \\ \Rightarrow (k+3)(k-5)=0 \end{gathered}$$

Keeping the value of factors equal to $0$,

$$\begin{gathered} \Rightarrow k+3=0 \\ \Rightarrow k=-3 \end{gathered}$$

Or,

$$\begin{gathered} \Rightarrow k-5=0 \\ \Rightarrow k=5 \end{gathered}$$

Therefore, the value of $k$ are $-3$ and $5$.