# How do you solve by completing the square, leaving answers in simplest radical form: x^2+ 3x- 2 = 0?

Apr 3, 2015

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Here is a solution for this question:

${x}^{2} + 3 x - 2 = 0$

$\left({x}^{2} + 3 x \textcolor{w h i t e}{\text{sssss}}\right) - 2 = 0$

!/2 of $3$ is $\frac{3}{2}$. Square that to get $\frac{9}{4}$ add $\frac{9}{4}$ to complete the square and subtract to keep the equation balanced:

$\left({x}^{2} + 3 x + \frac{9}{4} - \frac{9}{4}\right) - 2 = 0$ Regoup to keep just the perfect square in the parentheses:

$\left({x}^{2} + 3 x + \frac{9}{4}\right) - \frac{9}{4} - 2 = 0$

Factor the perfect square (use the $\frac{3}{2}$ from before and "$-$" like in the $+ 3 x$. Also simplify $- \frac{9}{2} - 2$

${\left(x + \frac{3}{2}\right)}^{2} - \frac{9}{4} - \frac{8}{4} = {\left(x + \frac{3}{2}\right)}^{2} - \frac{17}{4} = 0$

Solve ${\left(x + \frac{3}{2}\right)}^{2} - \frac{17}{4} = 0$ by "the Square Root Method"

${\left(x + \frac{3}{2}\right)}^{2} - \frac{17}{4} = 0$

${\left(x + \frac{3}{2}\right)}^{2} = \frac{17}{4}$

$x + \frac{3}{2} = \pm \sqrt{\frac{17}{4}} = \pm \frac{\sqrt{17}}{\sqrt{4}}$

$x + \frac{3}{2} = \pm \frac{\sqrt{17}}{2}$

$x = - \frac{3}{2} \pm \frac{\sqrt{17}}{2} = \frac{- 3 \pm \sqrt{17}}{2}$

(Use whichever form your teacher prefers -- one fraction or two.)