Question

If the roots of the equation $x^{2}-5 x+16=0$ are $\alpha, \beta$ and the roots of equation $x^{2}+p x+q=0$ 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 $x^{2}-5 x+16=0$ 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, $p=-1, q=-56$