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. $\gamma =\frac{\alpha }{4}+\beta$
• B. $\beta <\gamma$
• C. $\gamma =\frac{\alpha }{2}+\beta$
• D. $\alpha <\gamma <\beta$

Answer 1

Answer: (D)

$\alpha <\gamma <\beta$

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 $f\left(x\right)=x^2+6p^{2}x+10q^2$
Then,

$f\left(\alpha \right)={\alpha }^2+6p^2\alpha +10q^2$
$=({\alpha }^2+3p^2\alpha +5q^2)+3p^2\alpha +5q^2$
$=0+3p^2\alpha +5q^2$ [from equation (i)]
$\Rightarrow f\left(\alpha \right)>0$
and

$f\left(\beta \right)={\beta }^2+6p^2\beta +10q^2$
$=({\beta }^2+9p^2\beta +15q^2)-(3p^2\beta +5q^2)$
$=0-(3p^2\beta +5q^2)$ [from equation (ii)]
$\Rightarrow f\left(\beta \right)<0$

Thus, $f\left(x\right)$ is a polynomial such that $f\left(\alpha \right)>0$ and $f\left(\beta \right)<0$.
Therefore, there exists $\gamma$ satisfying $\alpha <\gamma <\beta$ such that $f\left(\gamma \right)=0$