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 $γ$ 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}+3 p^{2} x+5 q^{2}=0$
$\therefore \alpha^{2}+3 p^{2} \alpha+5 q^{2}=0\,\,\,...(i)$
and $\beta$ is a root of
$x^{2}+9 p^{2} x+15 q^{2}=0 $
$\therefore \beta^{2}+9 p^{2} \beta+15 q^{2}=0\,\,\,...(ii)$
Let $f(x)=x^{2}+6 p^{2} x+10 q^{2}$
Then, $ f(\alpha) =\alpha^{2}+6 p^{2} \alpha+10 q^{2} $
$=\left(\alpha^{2}+3 p^{2} \alpha+5 q^{2}\right)+3 p^{2} \alpha+5 q^{2} $
$=0+3 p^{2} \alpha+5 q^{2}\,\,\,$ [from Eq. (i)
$\Rightarrow f(\alpha) > 0$
and $f(\beta) =\beta^{2}+6 p^{2} \beta+10 q^{2} $
$=\left(\beta^{2}+9 p^{2} \beta+15 q^{2}\right)-\left(3 p^{2} \beta+5 q^{2}\right) $
$=0-\left(3 p^{2} \beta+5 q^{2}\right) \,\,\,$ [ 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 $f(\gamma)=0$.