Question

If $\alpha \text{and}\beta$ are the zeroes of $p\left(x\right)=4x^2+10x+k$ such that ${\alpha }^2+{\beta }^2+\alpha \beta =\frac{21}{4}$, then k =

• A. 2
• B. 4
• C. - 12
• D. 12

Answer 1

The correct option is B 4

Given, $\alpha \text{and}\beta$ are zeroes of a polynomial.
$\therefore \alpha +\beta =\frac{-10}{4}and\alpha \beta =\frac{k}{4}$

Now, it is given that
${\alpha }^2+{\beta }^2+\alpha \beta =\frac{21}{4}\Rightarrow (\alpha +\beta {)}^2-2\alpha \beta +\alpha \beta =\frac{21}{4}\Rightarrow {\left(\frac{-10}{4}\right)}^2-\alpha \beta =\frac{21}{4}\Rightarrow {\left(\frac{-10}{4}\right)}^2-\frac{k}{4}=\frac{21}{4}\Rightarrow \frac{100}{16}-\frac{k}{4}=\frac{21}{4}$
$\Rightarrow \frac{100-4k}{16}=\frac{21}{4}\Rightarrow 100-4k=84\Rightarrow 100-84=4k\Rightarrow k=4$