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

Jun 6, 2018

${x}_{1} = 3$, ${x}_{2} = - \frac{1}{2}$

#### Explanation:

Completing the square means that you want to write $2 {x}^{2} - 5 x - 3 = 0$
on the form ${\left(x - a\right)}^{2} = b$

First, let's get rid of the constant in front of the 2nd degree term:
${x}^{2} - \frac{5}{2} x - \frac{3}{2} = 0$
We want this on the form
${\left(x - a\right)}^{2} = b$
i.e. ${x}^{2} - 2 a x + {a}^{2} = b$
Comparing term for term, we see that
$2 a = \frac{5}{2}$, i.e. $a = \frac{5}{4}$
${\left(x - \frac{5}{4}\right)}^{2} = {x}^{2} - \frac{5}{2} x + \frac{25}{16}$
We, therefore, have
$\left({x}^{2} - \frac{5}{2} x + \frac{25}{16}\right) - \frac{25}{16} - \frac{3}{2} = 0$
or $\left({x}^{2} - \frac{5}{2} x + \frac{25}{16}\right) = \frac{25}{16} + \frac{24}{16} = \frac{49}{16}$
${\left(x - \frac{5}{4}\right)}^{2} = {\left(\frac{7}{4}\right)}^{2}$
$x - \frac{5}{4} = \pm \frac{7}{4}$
$x = \frac{5}{4} \pm \frac{7}{4}$
Therefore ${x}_{1} = \frac{5}{4} + \frac{7}{4} = 3$
${x}_{2} = \frac{5}{4} - \frac{7}{4} = - \frac{1}{2}$

Jun 6, 2018

$x = 3$ or $x = - \frac{1}{2}$

#### Explanation:

Given:

$2 {x}^{2} - 5 x - 3 = 0$

On dividing this equatin by $2$, we get:

${x}^{2} - \frac{5}{2} x - \frac{3}{2} = 0$

${x}^{2} - \frac{5}{2} x = \frac{3}{2}$

Add ${\left(\frac{5}{4}\right)}^{2}$ on both sides

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

[because ${a}^{2} + {b}^{2} + 2 a b = {\left(a + b\right)}^{2}$ ]

So

${\left(x - \frac{5}{4}\right)}^{2} = \frac{3}{2} + \frac{25}{16}$

${\left(x - \frac{5}{4}\right)}^{2} = \frac{49}{16}$

Taking the square root on both sides and solving it.

$x - \frac{5}{4} = \frac{7}{4} \text{ }$ or $\text{ } x - \frac{5}{4} = - \frac{7}{4}$

$x = \frac{7}{4} + \frac{5}{4} \text{ }$ or $\text{ } x = - \frac{7}{4} + \frac{5}{4}$

$x = \frac{12}{4} \text{ }$ or $\text{ } x = - \frac{2}{4}$

$x = 3 \text{ }$ or $\text{ } x = - \frac{1}{2}$