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 the value of p.

Answer 1

$x^2+px+15$

If $\alpha ,\beta$ be the roots

given ${(\alpha -\beta )}^2=196$

$(\alpha -\beta )=14\to 1$

$\alpha \beta =15$

${(\alpha +\beta )}^2={(\alpha -\beta )}^2+4\alpha \beta$

$=196+4\left(15\right)$

${(\alpha +\beta )}^2=256={16}^2$

$\alpha +\beta =16\to 2$

$\alpha +\beta =-p=16$

$\Rightarrow p=-16$