Question

Find the values of k for which the roots of the equation $(k+4)x^2+(k+1)x+1=0$ are real and equal?

Answer 1

$$\begin{gathered} \text{Given:} \\ (k+4)x^2+(k+1)x+1=0 \\ \text{Here,} \\ a=(k+4),b=(k+1)\mathrm{and}c=1 \\ \mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\text{the roots of the equation are real and equal; therefore, we have:} \\ D=0 \\ \Rightarrow b^2-4ac=0 \\ \Rightarrow {(k+1)}^2-4\times (k+4)\times 1=0 \\ \Rightarrow k^2+2k+1-4k-16=0 \\ \Rightarrow k^2-2k-15=0 \\ \Rightarrow k^2-5k+3k-15=0 \\ \Rightarrow k(k-5)+3(k-5)=0 \\ \Rightarrow (k-5)(k+3)=0 \\ \Rightarrow k=5\text{or}k=-3 \end{gathered}$$