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

Apr 25, 2016

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

#### Explanation:

We will use the difference of squares identity, which can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

First, to reduce the amount of arithmetic involving fractions, first multiply through by ${2}^{3} = 8$ to get:

$0 = 16 {x}^{2} + 40 x - 24$

$= {\left(4 x\right)}^{2} + 2 \left(5\right) \left(4 x\right) - 24$

$= {\left(4 x + 5\right)}^{2} - 25 - 24$

$= {\left(4 x + 5\right)}^{2} - {7}^{2}$

$= \left(\left(4 x + 5\right) - 7\right) \left(\left(4 x + 5\right) + 7\right)$

$= \left(4 x - 2\right) \left(4 x + 12\right)$

$= \left(2 \left(2 x - 1\right)\right) \left(4 \left(x + 3\right)\right)$

$= 8 \left(2 x - 1\right) \left(x + 3\right)$

Hence:

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

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Why did I premultiply by $8$?

One factor of $2$ makes the leading term into a perfect square, then the additional factor ${2}^{2} = 4$ compensates for the middle term being an odd number.

Otherwise, we might proceed as follows:

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

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

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

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

$= 2 \left({\left(x + \frac{5}{4}\right)}^{2} - {7}^{2} / {4}^{2}\right)$

$= 2 \left(\left(x + \frac{5}{4}\right) - \frac{7}{4}\right) \left(\left(x + \frac{5}{4}\right) + \frac{7}{4}\right)$

$= 2 \left(x - \frac{2}{4}\right) \left(x + \frac{12}{4}\right)$

$= 2 \left(x - \frac{1}{2}\right) \left(x + 3\right)$

Hence:

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

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Which method do you prefer?