80
$\alpha, \beta$ are the zeroes of the quadratic polynomial $p(x) = x^2 - 4x + k$, such that $\alpha - \beta = 8$. Find the value of $k$.
Show SolutionHide Solution↓
$p(x) = x^2 - 4x + k$
Here, $\alpha + \beta = 4$
and $\alpha \beta = k$
$(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta$
$\Rightarrow (8)^2 = (4)^2 - 4k$
$\Rightarrow k = -12$
Here, $\alpha + \beta = 4$
and $\alpha \beta = k$
$(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta$
$\Rightarrow (8)^2 = (4)^2 - 4k$
$\Rightarrow k = -12$