Question

If $\alpha$ and $\beta$ are the roots of the equation $x^2-8x+p=0$ such that ${\alpha }^2+{\beta }^2=40$, find the value of $p$.

• A. $8$
• B. $10$
• C. $12$
• D. $14$

Answer 1

The correct option is C $12$

As we know that,

$\alpha +\beta =\frac{-b}{a}$ and $\alpha \times \beta =\frac{c}{a}$

Hence,using the above equation we have

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

$40={\left(\frac{-b}{a}\right)}^2-2\left(\frac{p}{a}\right)$

$40=\frac{b^2-2ap}{a^2}$

Putting the values we get,

$40=\frac{8^2-2\times 1\times p}{1^2}$

$40=64-2p$

Hence , $p=12$