Question

If the square of the difference of the zeroes of the quadratic polynomial, $f\left(x\right)=x^2+px+15$ is equal to 196, then find out the value of p.

• A.
  $\pm 16$
• B.
  $\pm 17$
• C.
  $\pm 18$
• D.
  $\pm 19$

Answer 1

The correct option is A

$\pm 16$
$f\left(x\right)=x^2+px+15\text{Let zeroes of f(x) be}\alpha and\beta \alpha +\beta =\frac{-b}{a}=-p\alpha \beta =\frac{c}{a}=15(\alpha -\beta {)}^2=(\alpha +\beta {)}^2-4\alpha \beta \Rightarrow (\alpha -\beta {)}^2=(-p{)}^2-4\left(15\right)\Rightarrow (\alpha -\beta {)}^2=p^2-60But,(\alpha -\beta {)}^2=196\Rightarrow 196=p^2-60\Rightarrow 256=p^2\Rightarrow p=\pm 16$
So, option a is correct.