Question

Let $p$ and $q$be real numbers. If $\alpha$ is the root of $x^2+3p^{2}x+5q^2=0,\beta$ is the 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 root $\gamma$ that always satisfies.

• A. $ɣ=\left(\frac{ɑ}{4}\right)+\beta$
• B. $\beta <\gamma$
• C. $\gamma =\left(\frac{ɑ}{2}\right)+\beta$
• D. $\alpha <\beta <\gamma$

Answer 1

The correct option is D

$\alpha <\beta <\gamma$

Explanation for the correct option.

Finding root $\gamma$ always satisfies

Given: $\alpha$ is the root of $x^2+3p^{2}x+5q^2=0,\beta$ is the root of $x^2+9p^{2}x+15q^2=0$ on putting the values we get,

$\begin{array}{rcl}{x}^2+3p^{2}x+5q^2& =& 0,\\ & \Rightarrow & {\alpha }^2+3p^2\alpha +5q^2=0...\left(i\right)\\ & \Rightarrow & x^2+9p^{2}x+15q^2=0\end{array}$

${\beta }^2+9p^2\beta +15q^2=0...\left(ii\right)$

Now, let

$$\begin{gathered} f\left(x\right)=x^2+6p^{2}x+10q^2 \\ f\left(\alpha \right)={\alpha }^2+6p^2\alpha +10q^2 \\ =({\alpha }^2+3p^2\alpha +5q^2)+3p^2\alpha +5q^2 \\ =0+3p^2\alpha +5q^2 \\ \Rightarrow f\left(\alpha \right)>0 \end{gathered}$$

and again,

$$\begin{gathered} f\left(\beta \right)={\beta }^2+6p^2\beta +10q^2 \\ =({\beta }^2+9p^2\beta +15q^2)-(3p^2\beta +5q^2) \\ =0-(3p^2\beta +5q^2) \\ \Rightarrow f\left(\beta \right)<0 \end{gathered}$$

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 $f\left(\gamma \right)=0.$

Therefore, the correct answer is option (D).