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

1 Answer

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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
User Jorge Diaz
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