Question

lf the sum of the squares of the roots of $x^2+px-3=0$ is $10$, then $p=$

• A. $\pm 2$
• B. $\pm 3$
• C. $5$
• D. $-5$

Answer 1

The correct option is A $\pm 2$

Let $\alpha ,\beta$ are roots of $x^2+px-3=0,$ then

$S_1=\alpha +\beta =-p$

And $S_2=\alpha \beta =-3$

Given ${\alpha }^2+{\beta }^2=10$

Now from ${(\alpha +\beta )}^2={\alpha }^2+{\beta }^2+2\alpha \beta$

$\Rightarrow {(-p)}^2=10+2(-3)=10-6=4$

$\Rightarrow p^2=4$

$\Rightarrow p=\pm 2$