# How do I solve 25x^2 - 20x = 11 by completing the square?

Sep 13, 2014

Completing the square is method of solving a quadratic equation that involves finding a value to add to both the left and right side of the equation. This value has the extra benefit of making one side of the equation a perfect trinomial which makes the function easier to identify and/or graph.

Let's begin by factoring $25$ from the left side of the equation.

$25 \left({x}^{2} - \frac{20}{25} x\right) = 11$

Take the coefficient of the $x$ term and divide it by $2$ and square it

${\left(\frac{- \frac{20}{25}}{2}\right)}^{2} = {\left(- \frac{20}{25} \cdot \frac{1}{2}\right)}^{2} = {\left(- \frac{10}{25}\right)}^{2} = \frac{100}{625}$

$\frac{100}{625}$ This the number you add to the left side

$25 \left(\frac{100}{625}\right)$ Is added to the right side because we initially factored out 25 from the left side.

We added these values but the equation remains balanced because they are added to both sides of the equation.

$25 \left({x}^{2} - \frac{20}{25} x + \frac{100}{625}\right) = 11 + 25 \left(\frac{100}{625}\right)$

$25 \left({x}^{2} - \frac{20}{25} x + \frac{100}{625}\right) = 11 + \frac{100}{25}$

$25 \left({x}^{2} - \frac{20}{25} x + \frac{100}{625}\right) = 11 + 4$

$25 \left({x}^{2} - \frac{20}{25} x + \frac{100}{625}\right) = 15$

$\left({x}^{2} - \frac{20}{25} x + \frac{100}{625}\right) = \frac{15}{25}$

${\left(x - \frac{10}{25}\right)}^{2} = \frac{15}{25}$

$\sqrt{{\left(x - \frac{10}{25}\right)}^{2}} = \sqrt{\frac{15}{25}}$

$\left(x - \frac{10}{25}\right) = \sqrt{\frac{15}{25}}$

$x = \sqrt{\frac{15}{25}} + \frac{10}{25}$

$x = \frac{\sqrt{15}}{5} + \frac{10}{25}$

$x = \frac{\sqrt{15}}{5} + \frac{2}{5}$

$x = \frac{\sqrt{15} + 2}{5}$