How do you solve using the completing the square method x^2 = (3/4)x - (1/8)?

Mar 19, 2016

Multiply by $64$ first to cut down on the fractions, then complete the square and use the difference of squares identity to find:

$x = \frac{1}{2}$ or $x = \frac{1}{4}$

Explanation:

The difference of squares identity can be written:

${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right)$

I use this below, with $A = \left(8 x - 3\right)$ and $B = 1$.

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To match $a {x}^{2} + b x$ with a square, you normally look at:

$a {\left(x + \frac{b}{2 a}\right)}^{2} = a {x}^{2} + b x + {b}^{2} / \left(4 a\right)$

In our example, we can rearrange the original equation to get one involving $a = 1$ and $b = - \frac{3}{4}$, which would lead us to looking at:

${\left(x - \frac{3}{8}\right)}^{2} = {x}^{2} - \left(\frac{3}{4}\right) x + \frac{9}{64}$

These fractions get a little painful, so let us multiply the original equation by $64$ before we start and cut down on the fractions...

${x}^{2} = \left(\frac{3}{4}\right) x - \left(\frac{1}{8}\right)$

becomes:

$64 {x}^{2} = 48 x - 8$

which we can rearrange as:

$0 = 64 {x}^{2} - 48 x + 8$

$= {\left(8 x\right)}^{2} - 2 \left(8 x\right) \left(3\right) + 8$

$= {\left(8 x - 3\right)}^{2} - {3}^{2} + 8$

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

$= {\left(8 x - 3\right)}^{2} - 1$

$= {\left(8 x - 3\right)}^{2} - {1}^{2}$

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

$= \left(8 x - 4\right) \left(8 x - 2\right)$

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

$= 8 \left(2 x - 1\right) \left(4 x - 1\right)$

So $x = \frac{1}{2}$ or $x = \frac{1}{4}$