How do you find a quadratic polynomial with integer coefficients which has x=3/5+-sqrt29/5 as its real zeros?

Sep 21, 2017

$5 {x}^{2} - 6 x - 4$

Explanation:

$\text{given the zeros ( roots) of a quadratic } x = a , x = b$

$\text{then the factors are } \left(x - a\right) , \left(x - b\right)$

$\text{and the quadratic is the product of the factors}$

$\Rightarrow p \left(x\right) = \left(x - a\right) \left(x - b\right)$

$\text{here the roots are "x=3/5+sqrt29/5" and } x = \frac{3}{5} - \frac{\sqrt{29}}{5}$

$\Rightarrow \text{factors } \left(x - \left(\frac{3}{5} + \frac{\sqrt{29}}{5}\right)\right) , \left(x - \left(\frac{3}{5} - \frac{\sqrt{29}}{5}\right)\right)$

$\Rightarrow p \left(x\right) = \left(x - \frac{3}{5} - \frac{\sqrt{29}}{5}\right) \left(x - \frac{3}{5} + \frac{\sqrt{29}}{5}\right)$

$\textcolor{w h i t e}{\Rightarrow p \left(x\right)} = {\left(x - \frac{3}{5}\right)}^{2} - {\left(\frac{\sqrt{29}}{5}\right)}^{2}$

$\textcolor{w h i t e}{\Rightarrow p \left(x\right)} = {x}^{2} - \frac{6}{5} x + \frac{9}{25} - \frac{29}{25}$

$\textcolor{w h i t e}{\Rightarrow p \left(x\right)} = {x}^{2} - \frac{6}{5} x - \frac{4}{5} \to \left(\times 5\right)$

$\textcolor{w h i t e}{\Rightarrow p \left(x\right)} = 5 {x}^{2} - 6 x - 4$

Sep 21, 2017

$5 {x}^{2} - 6 x - 4 = 0$

Explanation:

if $\alpha \text{ & "beta" are the roots of a quadratic equation}$

the equation can be written as

${x}^{2} - \left(\alpha + \beta\right) x + \alpha \beta = 0$

we have

$\alpha = \frac{3}{5} + \frac{\sqrt{29}}{5}$

$\beta = \frac{3}{5} - \frac{\sqrt{29}}{5}$

$\alpha + \beta = \frac{3}{5} + \cancel{\frac{\sqrt{29}}{5}} + \frac{3}{5} - \cancel{\frac{\sqrt{29}}{5}}$

$\alpha + \beta = \frac{6}{5}$

$\alpha \beta = \left(\frac{3}{5} + \frac{\sqrt{29}}{5}\right) \left(\frac{3}{5} - \frac{\sqrt{29}}{5}\right)$

$= \frac{9}{25} - \frac{29}{25} = - \frac{20}{25} = - \frac{4}{5}$

$\therefore {x}^{2} - \left(\alpha + \beta\right) x + \alpha \beta$

becomes

${x}^{2} - \frac{6}{5} x - \frac{4}{5} = 0$

$\implies 5 {x}^{2} - 6 x - 4 = 0$