Question

Let $p,q$ be real numbers. If $\alpha$ is the root of $x^2+3p^{2}x+5q^2=0,\beta$ is a root of $x^2+9p^{2}x+15q^2=0$ and $0<\alpha <\beta ,$ then the equation $x^2+6p^{2}x+10q^2=0$ has a root $\gamma$ that always satisfies

• A. $\gamma =\alpha /4+\beta$
• B. $\beta <\gamma$
• C. $\gamma =\alpha /2+\beta$
• D. $\alpha <\gamma <\beta$

Answer 1

Answer: (D)

Since, $\alpha$ is a root of
$x^2+3p^{2}x+5q^2=0$
$\therefore {\alpha }^2+3p^2\alpha +5q^2=0...(i)$
and $\beta$ is a root of
$x^2+9p^{2}x+15q^2=0$
$\therefore {\beta }^2+9p^2\beta +15q^2=0...(ii)$
Let $f(x)=x^2+6p^{2}x+10q^2$
Then, $f(\alpha )={\alpha }^2+6p^2\alpha +10q^2$
$=\left({\alpha }^2+3p^2\alpha +5q^2\right)+3p^2\alpha +5q^2$
$=0+3p^2\alpha +5q^2$ [from Eq. (i)
$\Rightarrow f(\alpha )>0$
and $f(\beta )={\beta }^2+6p^2\beta +10q^2$
$=\left({\beta }^2+9p^2\beta +15q^2\right)-\left(3p^2\beta +5q^2\right)$
$=0-\left(3p^2\beta +5q^2\right)$ [ from Eq. (ii)]
$\Rightarrow f(\beta )<0$
Thus, $f(x)$ is a polynomial such that $f(\alpha )>0$ and $f(\beta )<0.$
Therefore, there exists $\gamma$ satisfying $\alpha <\gamma <\beta$ such that $f(\gamma )=0$.