Question

If the sum of the squares of the roots of $x^2-(p-2)x-(p+1)=0\left(p\epsilon R\right)$ is 5, then what is the value of p?

• A. -1
• B. 1
• C. $\frac{3}{2}$

Answer 1

The correct option is B 1

Let $\alpha$ and $\beta$ be the roots of $x^2-(p-2)x-(p+1)=0$
Then, $\alpha +\beta =p-2$
and $\alpha \beta =-(p+1)\therefore {\alpha }^2+{\beta }^2=5\Rightarrow (\alpha +\beta {)}^2-2\alpha \beta =5\Rightarrow (p-2{)}^2+2(p+1)=5\Rightarrow p^2-4p+4+2p+2=5\Rightarrow p^2-2p+1=0\Rightarrow (p-1{)}^2=0\Rightarrow p=1$