Question

If $\alpha ,\beta$ are the zeroes of the polynomial $x^2-px+36$ and ${\alpha }^2$+${\beta }^2$ = 9, then what is the value of p?

• A. ±6
• B. ±3
• C. ±8
• D. ±9

Answer 1

The correct option is D

±9

Given polynomial $x^2-px+36$

On comparing with the standard form of a quadratic polynomial $a{x}^2+bx+c$, we get
a = 1, b = -p, c = 36

Here, $\alpha$ and $\beta$ are the zeroes of the polynomial.
$\Rightarrow \alpha +\beta =\frac{-b}{a}=p$
and $\alpha \beta =\frac{c}{a}=36$

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

$\Rightarrow$ $9=p^2-2\times 36$ [$\because {\alpha }^2+{\beta }^2$ = 9]
$\Rightarrow 81=p^2$

$\Rightarrow p=9\text{or}-9$