Question

If $\alpha ,\beta$ are the real and distinct roots of $x^2+px+q=0$ and ${\alpha }^4,{\beta }^4$ are the roots of $x^2-rx+5=0$, then the equation $x^2-4qx+2a^2-r=0$ always has:

• A. two real roots
• B. one positive root and one negative root
• C. two positive roots
• D. two negative roots

Answer 1

Answer: (A), (B)

one positive root and one negative root

$\alpha ,\beta$ are the roots of $x^2+px+q=0$
$\Rightarrow \alpha +\beta =-p$ and $\alpha \beta =q$
${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta =p^2-2q$
${\alpha }^4+{\beta }^4=({\alpha }^2+{\beta }^2{)}^2-2{\alpha }^2{\beta }^2$
$=(p^2-2q{)}^2-2q^2$
${\alpha }^4$ and ${\beta }^4$ are the roots of $x^2-rx+s=0$
$\Rightarrow {\alpha }^4{\beta }^4=s$
${\alpha }^4+{\beta }^4=r$

we Know ${\alpha }^4+{\beta }^4=(p^2-2q{)}^2-2q^2$
$\Rightarrow (p^2-2q{)}^2-2q^2=r$ .............(1)
The roots of the equation $x^2-4qx+2q^2-r=0$ are

$x=\frac{4q\pm \sqrt{(4q{)}^2-4(2q^2-r)}}{2}$
$x=\frac{4q\pm 2\sqrt{4q^2-2q^2+r}}{2}$
$x=2q\pm \sqrt{2q^2+r}$

From (1), $2q^2+r=(p^2-2q{)}^2$
$\Rightarrow$ Roots $=2q\pm \sqrt{(p^2-2q{)}^2}$
$=2q\pm (p^2-2q)$
One of the roots = $2q+(p^2-2q)$
$\Rightarrow x=p^2$, which is +ve
Second root is $2q-(p^2-2q)$
$\Rightarrow x=4q-p^2$
$p^2-4q>0$, since the first quadratic has two real roots $\Rightarrow 4q-p^2$ is negative

$\underset{–}{Method2}$
For the equation $x^2+px+q,\alpha ,\beta$ are the roots
$\Rightarrow (\alpha +\beta )=-p$ and $\alpha \beta =q$
${\alpha }^4+{\beta }^4=(\alpha +\beta {)}^2-2\alpha \beta =p^2-2q$
${\alpha }^4+{\beta }^4=({\alpha }^2+{\beta }^2{)}^2-2{\alpha }^2{\beta }^2=p^2-2q{)}^2-2q^2$ ...........(1)

Now, for the equation $x^2-rx+s=0,{\alpha }^4,{\beta }^4$ are the roots.
$\Rightarrow {\alpha }^4{\beta }^4=s$ and ${\alpha }^4+{\beta }^4=r$ .........(2)
From(1), (2)
$r=(p^2-2q{)}^2-2q^2$ ..........(3)
Consider the equation $x^2-4ax+2q^2-r=0$,
Product of the roots $=2q^2-r$
$=2q^2-\left[\right(p^2-2q{)}^2-2q^2]$ (using(3))
$=2q^2-[p^4+4q^2-4q{p}^2-2q^2]$
$=-p^4+4q{p}^2$

$(p^2-4q>0$ [ 1st equation has real roots ]
Product of roots = -ve
$\Rightarrow$ Roots are of opposite sign, if they are real
Consider discriminant of the equation

$x^2-4ax+2q^2-r=0$
$△=\left(4q^2\right)-4(2q^2-r)$
$=16q^2-4(2q^2-r)$
$=4(4q^2-2q^2+r)$
$=4(2q^2+r)$

$r+2q^2=(p^2-2q{)}^2(from\left(3\right))$
$\Rightarrow △$ is +ve
So, the roots are real and of opposite sign