Question

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

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

Answer 1

Answer: (B)

$p=-1$ and $q=-56$

Given:
First quadratic equation: $x^2-5x+16=0$ and its roots are $\alpha$ and $\beta$;
Second quadratic equation: $x^2+px+q=0$ and its roots are $({\alpha }^2+{\beta }^2)$ and $\frac{\alpha \beta }{2}$.
We know that the standard quadratic equation is: $a{x}^2+bx+c=0$
Comparing the first equation with the standard equation, we get $a=1,b=-5$ and $c=16$.
We also know that sum of the roots $(\alpha +\beta )=-\frac{b}{a}=-\frac{(-5)}{1}=5$.
And product of the roots $\left(\alpha \beta \right)=\frac{c}{a}=\frac{16}{1}=16$.
We also know that ${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta =-7$
Since $({\alpha }^2+{\beta }^2)$ and $\frac{\alpha \beta }{2}$ are roots of equation,
$x^2+px+q=0$,$({\alpha }^2+{\beta }^2)+\frac{\alpha \beta }{2}=-p\Rightarrow p=-1$
$({\alpha }^2+{\beta }^2)\left(\frac{\alpha \beta }{2}\right)=q\Rightarrow q=-56$.
Hence, option 'B' is correct.