If the equation whose roots are p times the r | vedclass

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(B) Let the roots of the given equation $f(x) = x^4-2ax^3+4bx^2+8ax+16=0$ be $\alpha_1, \alpha_2, \alpha_3, \alpha_4$.The roots of the new equation ar

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$\frac{1}{2}$

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(B) Let the roots of the given equation $f(x) = x^4-2ax^3+4bx^2+8ax+16=0$ be $\alpha_1, \alpha_2, \alpha_3, \alpha_4$.The roots of the new equation are $p\alpha_1, p\alpha_2, p\alpha_3, p\alpha_4$.Thus,the new equation is $f(\frac{x}{p}) = 0$.Substituting $\frac{x}{p}$ into the equation:$(\frac{x}{p})^4 - 2a(\frac{x}{p})^3 + 4b(\frac{x}{p})^2 + 8a(\frac{x}{p}) + 16 = 0$Multiplying by $p^4$:$x^4 - 2apx^3 + 4bp^2x^2 + 8ap^3x + 16p^4 = 0$Since this is a reciprocal equation,the coefficient of $x^4$ must equal the constant term:$1 = 16p^4$$p^4 = \frac{1}{16}$$p^2 = \frac{1}{4} \implies |p| = \frac{1}{2}$

If the equation whose roots are $p$ times the roots of the equation $x^4-2ax^3+4bx^2+8ax+16=0$ is a reciprocal equation,then $|p|=$ :

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