Question

If the roots of the equation $x^2-(p+4)x+2p+5=0$ are real and equal, then the value(s) of $p$ is/are

• A. $2$
• B. $-2$
• C. $-1$
• D. $1$

Answer 1

Answer: (A), (B)

$-2$

Given: $x^2-(p+4)x+2p+5=0$
The roots are real and equal $\Rightarrow D=0\Rightarrow \{-(p+4){\}}^2-4(2p+5)=0\Rightarrow (p+4{)}^2-8p-20=0\Rightarrow p^2+8p+16-8p-20=0\Rightarrow p^2-4=0\Rightarrow p=\pm 2$
Hence, $p=\pm 2$ are the two possible cases for which the roots of the equation are real and equal.