Question

The roots of 2x^{2} + 3x = 4 are α and β. Find the simplest quadratic equation which has roots =%5Cfrac%7B1%7D%7Balpha%7D%20" id="TexFormula1" title="\frac{1}{alpha} " alt="\frac{1}{alpha} " align="absmiddle" class="latex-formula"> and $\frac{1}{beta}$.

Answer 1

$2x^2+3x=4\implies2x^2+3x-4=0\implies x=\dfrac{-3\pm\sqrt{41}}4$

Let $\alpha$ be the root with the positive square root and $\beta$ the root with the negative square root. Then

$\dfrac1\alpha=\dfrac4{-3+\sqrt{41}}\quad\text{and}\quad\dfrac1\beta=\dfrac4{-3-\sqrt{41}}$

The simplest quadratic with these roots can be written as

$\left(x-\dfrac1\alpha\right)\left(x-\dfrac1\beta\right)=x^2-\left(\dfrac1\alpha+\dfrac1\beta\right)x+\dfrac1{\alpha\beta}=x^2-\dfrac34x-\dfrac12$