Question

Suppose ${{x}^{2}}+px+q=0$ has two real roots $\alpha $ and $\beta $ with $\left| \alpha \right|\ne \left| \beta \right|$. If ${{\alpha }^{4}}$ and ${{\beta }^{4}}$ are the root of ${{x}^{2}}+rx+s=0$, then the equation ${{x}^{2}}-4qx+2{{q}^{2}}+r=0$ has
A. one positive and one negative root.
B. Two distinct positive roots.
C. two distinct negative roots.
D. no real roots.

Answer 1

Hint: The given problem is to determine the nature of the roots of the equation ${{x}^{2}}-4qx+2{{q}^{2}}+r=0$. We know that the nature of the roots of a quadratic equation $ax+by+c=0$ are depends on the value of ${{b}^{2}}-4ac$.
If
${{b}^{2}}-4ac>0$, the equation has two roots and they are real and distinct.
${{b}^{2}}-4ac=0$, the equation has only one real root.
${{b}^{2}}-4ac<0$, the equation has two distinct imaginary roots.
In the problem we have the $\alpha $ and $\beta $ are the roots of ${{x}^{2}}+px+q=0$, ${{\alpha }^{4}}$ and ${{\beta }^{4}}$ are the roots of ${{x}^{2}}+rx+s=0$ from this we list out the equations by using the relation between the roots and coefficients of the quadratic equation and calculates the value of ${{b}^{2}}-4ac$ for the equation ${{x}^{2}}-4qx+2{{q}^{2}}+r=0$.

Complete step-by-step answer:
Given that,
$\alpha $ and $\beta $ are the roots of ${{x}^{2}}+px+q=0$
We can write $\alpha +\beta =-p$, $\alpha \beta =q$
${{\alpha }^{4}}$ and ${{\beta }^{4}}$ are the roots of ${{x}^{2}}+rx+s=0$
We can write ${{\alpha }^{4}}+{{\beta }^{4}}=-r$, ${{\alpha }^{4}}.{{\beta }^{4}}=s$.
Given equation ${{x}^{2}}-4qx+2{{q}^{2}}+r=0$, now the value of ${{b}^{2}}-4ac$ for the above equation is given by
$\begin{align}
  & {{b}^{2}}-4ac={{\left( -4q \right)}^{2}}-4\left( 1 \right)\left( 2{{q}^{2}}+r \right) \\
 & \Rightarrow {{b}^{2}}-4ac=16{{q}^{2}}-8{{q}^{2}}-4r \\
 & \Rightarrow {{b}^{2}}-4ac=8{{q}^{2}}-4r....\left( \text{i} \right) \\
\end{align}$
We have $\alpha \beta =q$ and ${{\alpha }^{4}}+{{\beta }^{4}}=-r$ substituting these values in the above equation, then we will get
$\begin{align}
  & {{b}^{2}}-4ac=8{{\left( \alpha \beta \right)}^{2}}+4\left( {{\alpha }^{4}}+{{\beta }^{4}} \right) \\
 & \Rightarrow {{b}^{2}}-4ac=8{{\alpha }^{2}}{{\beta }^{2}}+4{{\alpha }^{4}}+4{{\beta }^{4}} \\
 & \Rightarrow {{b}^{2}}-4ac={{\left( 2{{\alpha }^{2}} \right)}^{2}}+{{\left( 2{{\alpha }^{2}} \right)}^{2}}+2.\left( 2{{\alpha }^{2}} \right)\left( 2{{\beta }^{2}} \right) \\
 & \Rightarrow {{b}^{2}}-4ac={{\left( 2{{\alpha }^{2}}+2{{\beta }^{2}} \right)}^{2}}>0 \\
\end{align}$
So, the equation ${{x}^{2}}-4qx+2{{q}^{2}}+r=0$ always has two distinct positive roots.

So, the correct answer is “Option B”.

Note: Sometimes in the problem they may also ask to find the roots of the equation ${{x}^{2}}-4qx+2{{q}^{2}}+r=0$, then we will use the known formula roots of the quadratic equation $ax+by+c=0$ are $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ and calculates the roots of the equations in terms of $\alpha $ and $\beta $.