Question

if 1/root alpha and 1/ roots beta are root of equation ax square + bx+1 =0 then prove that the equation x (x+b cube ) +(a cube _ 3bx )=0 has roots alpha to the power 3 upon2 , beta to the power 3 upon2​

Answer 1

$\textbf{Given:}$

$\text{ \dfrac{1}{\sqrt{\alpha}} and \dfrac{1}{\sqrt{\beta}} are roots of ax^2+bx+1=0 }$

$\textbf{To prove:}$

$\text{ {\alpha}^\frac{3}{2} and {\beta}^\frac{3}{2} are roots of \bf\,x^2+(b^3-3ab)x+a^3=0 }$

$\textbf{Solution:}$

$\text{Since \dfrac{1}{\sqrt{\alpha}} and \dfrac{1}{\sqrt{\beta}} are roots of ax^2+bx+1=0 ,}$

$\text{we have}$

$\dfrac{1}{\sqrt{\alpha}}+\dfrac{1}{\sqrt{\beta}}=\dfrac{-b}{a}$.......(1)

$\dfrac{1}{\sqrt{\alpha}}\dfrac{1}{\sqrt{\beta}}=\dfrac{1}{a}$.......(2)

$\implies\dfrac{1}{\sqrt{\alpha\,\beta}}=\dfrac{1}{a}$

$\implies\boxed{\sqrt{\alpha\,\beta}=a}$

$\text{From (1),}$

$\dfrac{\sqrt{\alpha}+\sqrt{\beta}}{\sqrt{\alpha\,\beta}}=\dfrac{-b}{a}$

$\dfrac{\sqrt{\alpha}+\sqrt{\beta}}{a}=\dfrac{-b}{a}$

$\implies\boxed{\sqrt{\alpha}+\sqrt{\beta}=-b}$

$\text{Now, we form a quadratic equation whose roots are}$

$\text{ {\alpha}^\frac{3}{2} and {\beta}^\frac{3}{2} }$

$\textbf{Sum of the roots}$

$={\alpha}^\frac{3}{2}+{\beta}^\frac{3}{2}$

$=({\alpha}^\frac{1}{2})^3+({\beta}^\frac{1}{2})^3$

$=({\alpha}^\frac{1}{2}+{\beta}^\frac{1}{2})(\alpha-{\alpha}^\frac{1}{2}{\beta}^\frac{1}{2}+\beta)$

$=(\sqrt{\alpha}+\sqrt{\beta})(\alpha+\beta-\sqrt{\alpha\,\beta})$

$=(\sqrt{\alpha}+\sqrt{\beta})((\sqrt{\alpha}+\sqrt{\beta})^2-2\sqrt{\alpha\,\beta}-\sqrt{\alpha\,\beta})$

$=(\sqrt{\alpha}+\sqrt{\beta})((\sqrt{\alpha}+\sqrt{\beta})^2-3\sqrt{\alpha\,\beta})$

$=(-b)((-b)^2-3a)$

$=(-b)(b^2-3a)$

$=-b^3+3ab$

$\textbf{Product of the roots}$

$={\alpha}^\frac{3}{2}\,{\beta}^\frac{3}{2}$

$=({\alpha}^\frac{1}{2}\,{\beta}^\frac{1}{2})^3$

$=(\sqrt{\alpha}\sqrt{\beta})^3$

$=(\sqrt{\alpha\,\beta})^3$

$=a^3$

$\textbf{The required quadratic equation is}$

$x^2-(-b^3+3ab)x+a^3=0$

$\boxed{\bf\,x^2+(b^3-3ab)x+a^3=0}$

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