Question

If $\alpha \text{and}\beta$ are the zeroes of $p\left(x\right)=8x^2+5x+k$ such that ${\alpha }^2+{\beta }^2+\alpha \beta =\frac{15}{4}$, then k =

Answer 1

Given, $\alpha \text{and}\beta$ are zeroes of a polynomial.
$\therefore \alpha +\beta =\frac{-5}{8}and\alpha \beta =\frac{k}{8}$

Now, it is given that
${\alpha }^2+{\beta }^2+\alpha \beta =\frac{15}{4}\left(0.5marks\right)\Rightarrow (\alpha +\beta {)}^2-2\alpha \beta +\alpha \beta =\frac{15}{4}\Rightarrow {\left(\frac{-5}{8}\right)}^2-\alpha \beta =\frac{15}{4}\Rightarrow {\left(\frac{-5}{8}\right)}^2-\frac{k}{8}=\frac{15}{4}\Rightarrow \frac{25}{64}-\frac{k}{8}=\frac{15}{4}$
$\Rightarrow \frac{25-8k}{64}=\frac{15}{4}\left(0.5marks\right)\Rightarrow 25-8k=\frac{64\times 15}{4}\Rightarrow 25-8k=\frac{960}{4}\Rightarrow 25-8k=240\Rightarrow 8k=25-240$
8k = -215
k = $\frac{-215}{8}$.

(1 mark)