Question

If the roots of the equation $x^2-5x+16=0$ are $\alpha ,\beta$ and the roots of equation $x^2+px+q=0$ are ${\alpha }^2+{\beta }^2,\frac{\alpha \beta }{2}$, then

• A. p = 1, q = -56
• B. p = -1, q = -56
• C. p = 1, q = 56
• D. p = -1, q = 56

Answer 1

Answer: (B)

Since roots of the equation $x^2-5x+16=0$ are $\alpha ,\beta$.
$\Rightarrow \alpha +\beta =5$ and $\alpha \beta =16$ and ${\alpha }^2+{\beta }^2+\frac{\alpha \beta }{2}=-p$
$\Rightarrow (\alpha +\beta {)}^2-2\alpha \beta +\frac{\alpha \beta }{2}=-p$ $\Rightarrow 25-32+8=-p$
$\Rightarrow p=-1$ and $\left({\alpha }^2+{\beta }^2\right)\left(\frac{\alpha \beta }{2}\right)=q$
$\Rightarrow \left[(\alpha +\beta {)}^2-2\alpha \beta \right]\left[\frac{\alpha \beta }{2}\right]=q$
$\Rightarrow q=[25-32]\frac{16}{2}=-56$
So, $p=-1,q=-56$