Answer to Question #236228 in Math for luka

Question

The roots of the equation 25X2+ X +1 =0 are α^2 and β^2. Find the equation with

integral coefficients whose roots are 1/α and 1/β

Answer 1

$25x^2+x+1=0$

By Viet Theorem

$\alpha^2+\beta^2=-\dfrac{1}{25}$

$\alpha^2\beta^2=\dfrac{1}{25}$

Then

$\dfrac{1}{\alpha^2}+\dfrac{1}{\beta^2}=\dfrac{\alpha^2+\beta^2}{\alpha^2\beta^2}$

$=\dfrac{-\dfrac{1}{25}}{\dfrac{1}{25}}=-1$

$\alpha\beta=-\dfrac{1}{5}\ \text{or}\ \alpha\beta=\dfrac{1}{5}$

$\alpha\beta=-\dfrac{1}{5}$

$\dfrac{1}{\alpha}(\dfrac{1}{\beta})=-5$

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

$=-1+2(-5)=-11$

$\dfrac{1}{\alpha}+\dfrac{1}{\beta}=-i\sqrt{11}\text{ or }\dfrac{1}{\alpha}+\dfrac{1}{\beta}=i\sqrt{11}$

The quadratic equation is

$y^2+i\sqrt{11}y-5=0$

Or

$y^2-i\sqrt{11}y-5=0$

$\alpha\beta=\dfrac{1}{5}$

$\dfrac{1}{\alpha}(\dfrac{1}{\beta})=5$

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

$=-1+2(5)=9$

$\dfrac{1}{\alpha}+\dfrac{1}{\beta}=-3\text{ or }\dfrac{1}{\alpha}+\dfrac{1}{\beta}=3$

The quadratic equation is

$y^2+3y+5=0$

Or

$y^2-3y+5=0$

$y^2+i\sqrt{11}y-5=0$

$y^2-i\sqrt{11}y-5=0$

$y^2+3y+5=0$

$y^2-3y+5=0$