# How do you solve the equation 2x^2+3x-5=0 by completing the square?

Dec 21, 2016

x=-5/2 " and x=1

#### Explanation:

Write as $2 \left({x}^{2} + \frac{3}{2} x\right) - 5 + k = 0. \ldots \ldots E q u a t i o n \left(1\right)$

Where $k$ is a constant of correction that compensates for the error introduce whilst manipulating the equation.

Take the power outside the bracket.

$2 {\left(x + \frac{3}{2} x\right)}^{2} - 5 + k = 0$

Remove the $x$ from $\frac{3}{2} x$

$2 {\left(x + \frac{3}{2}\right)}^{2} - 5 + k = 0$

Halve the $\frac{3}{2}$

$2 {\left(x + \frac{3}{4}\right)}^{2} - 5 + k = 0$

$\textcolor{red}{\text{The error comes from}}$

$\textcolor{red}{2} {\left(x + \textcolor{red}{\frac{3}{4}}\right)}^{\textcolor{red}{2}} - 5 + k = 0. . . E q u a t i o n \left({1}_{a}\right)$
.................................................................................................................
in that we have $\textcolor{red}{2 \times {\left(\frac{3}{4}\right)}^{2}} + k = 0 \leftarrow \text{ building the error correction}$

$\implies k = - \left(2 \times \frac{9}{16}\right) = - \frac{18}{16} = - \frac{9}{8}$
.....................................................................................................................

$\implies 2 {\left(x + \frac{3}{4}\right)}^{2} - 5 - \frac{9}{8} = 0$

$\implies \textcolor{b l u e}{2 {\left(x + \frac{3}{4}\right)}^{2} - \frac{49}{8} = 0} \ldots \ldots \ldots \ldots \ldots \ldots . . E q u a t i o n \left({1}_{b}\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

${\left(x + \frac{3}{4}\right)}^{2} = \frac{49}{16}$

$x + \frac{3}{4} = \pm \sqrt{\frac{49}{16}} \text{ "=" } \pm \frac{7}{4}$

$x = \pm \frac{7}{4} - \frac{3}{4}$

$x = - \frac{10}{4} \text{ and } 1$

$\textcolor{red}{x = - \frac{5}{2} \text{ and } 1}$