Question

If the roots of quadratic equation $(2k+3)x^2+2(k+3)x+(k+5)=0,where\left(k\in R,k\ne \frac{-3}{2}\right)$ are equal, the value of $k$

• A. $1,6$
• B. $–1,–6$
• C. $–1,6$
• D. $1,–6$

Answer 1

The correct option is B

$–1,–6$

Step 1: Find the Discriminant D of the above equation $D=b^2-4ac$

For this, we compare $(2k+3)x^2+2(k+3)x+(k+5)=0$ with the standard form $a{x}^2+bx+c$

Here,

$$\begin{gathered} b=2(k+3) \\ a=2k+3 \\ c=k+5 \\ \therefore D=b^2-4ac \\ =(2{(k+3))}^2-4(2k+3\left)\right(k+5) \\ =4(k^2+9+6k)-4(2k^2+10k+3k+15) \\ =4k^2+\hspace{0.17em}36+24k-8k^2-40k-12k-60 \\ =-4k^2-28k-24 \end{gathered}$$

Step 2: Set Discriminant $D=0$

$$\begin{gathered} \Rightarrow -4k^2-28k-24=0 \\ \Rightarrow k^2+7k+6=0[Dividingbothsidesby-4] \\ \Rightarrow k^2+6k+k+6=0[Factorizingbymidtermsplitting] \\ \Rightarrow k(k+6)+1(k+6)=0 \\ \Rightarrow (k+6)(k+1)=0 \\ \Rightarrow (k+6)=0and(k+1)=0 \\ \Rightarrow k=-6andk=-1 \\ \mathrm{Hence},k=-6,-1 \end{gathered}$$

Therefore, option (2) is correct.