Question

The roots of 2x^{2} + 3x = 4 are α and β. Find the simplest quadratic equation which has roots
`(1)/(alpha)` and
`(1)/(beta)`.

Answer 1

`2x^2+3x=4\implies2x^2+3x-4=0\implies x=\frac{-3\pm√(41)}4`

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


`\frac1\alpha=\frac4{-3+√(41)}\quad\text{and}\quad\frac1\beta=\frac4{-3-√(41)}`

The simplest quadratic with these roots can be written as


`\left(x-\frac1\alpha\right)\left(x-\frac1\beta\right)=x^2-\left(\frac1\alpha+\frac1\beta\right)x+\frac1{\alpha\beta}=x^2-\frac34x-\frac12`