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If the equation whose roots are p times the roots of the equation $x^4-2 a x^3+4 b x^2+8 a x+16=0$ is a reciprocal equation, then $|\mathrm{p}|=$ :
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Verified Answer
The correct answer is:
$\frac{1}{2}$
$\because$ Roots of other equation is $P$ times the root of given equation.
$$
\begin{aligned}
& \therefore f\left(\frac{x}{P}\right)=0 \\
& \left(\frac{x}{P}\right)^4-2 a\left(\frac{x}{P}\right)^3+4 b\left(\frac{x}{P}\right)^2+8 a\left(\frac{x}{P}\right)+16=0 \\
& \Rightarrow x^4-2 a P x^3+4 b P^2 x^2+8 a P^3 x+16 P^4=0
\end{aligned}
$$
$\because$ It is reciprocal equation
$$
\begin{aligned}
& \therefore 16 P^4=\text { Coeff. of } x^4 \text { in } f(x) \\
& \Rightarrow 16 P^4=1 \\
& \Rightarrow P^4=\frac{1}{16} \Rightarrow P= \pm \frac{1}{2} \Rightarrow|P|=\frac{1}{2}
\end{aligned}
$$
$$
\begin{aligned}
& \therefore f\left(\frac{x}{P}\right)=0 \\
& \left(\frac{x}{P}\right)^4-2 a\left(\frac{x}{P}\right)^3+4 b\left(\frac{x}{P}\right)^2+8 a\left(\frac{x}{P}\right)+16=0 \\
& \Rightarrow x^4-2 a P x^3+4 b P^2 x^2+8 a P^3 x+16 P^4=0
\end{aligned}
$$
$\because$ It is reciprocal equation
$$
\begin{aligned}
& \therefore 16 P^4=\text { Coeff. of } x^4 \text { in } f(x) \\
& \Rightarrow 16 P^4=1 \\
& \Rightarrow P^4=\frac{1}{16} \Rightarrow P= \pm \frac{1}{2} \Rightarrow|P|=\frac{1}{2}
\end{aligned}
$$
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