Question

If the sum of the squares of the zeroes of the quadratic polynomial, $f\left(x\right)=x^2-px+70$ is 149, then the value of p is

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

Answer 1

The correct option is B

$\pm 17$
$f\left(x\right)=x^2-px+70\text{Let zeroes of f(x) be}\alpha and\beta {\alpha }^2+{\beta }^2=149Now,\alpha +\beta =\frac{-b}{a}=\frac{-(-p)}{1}=P\alpha \beta =\frac{c}{a}=\frac{70}{1}=70{\alpha }^2+{\beta }^2=149\Rightarrow ({\alpha }^2+{\beta }^2+2\alpha \beta )-2\alpha \beta =149\Rightarrow (\alpha +\beta {)}^2-2\alpha \beta =149\Rightarrow (p{)}^2-2\left(70\right)=149\Rightarrow p^2-140=149\Rightarrow p^2=149+140\Rightarrow p^2=289\Rightarrow p=\pm 17$
So, option b is correct.