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

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

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Answer 1 · 2016-03-19T09:09:12

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

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

Explanation:

The difference of squares identity can be written:

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

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

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

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

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

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

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

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

becomes:

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

which we can rearrange as:

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

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

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

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

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

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

$= ((8 x - 3) - 1) ((8 x - 3) + 1)$ $= ((8x-3)-1)((8x-3)+1)$

$= (8 x - 4) (8 x - 2)$ $= (8x-4)(8x-2)$

$= (4 (2 x - 1)) (2 (4 x - 1))$ $= (4(2x-1))(2(4x-1))$

$= 8 (2 x - 1) (4 x - 1)$ $= 8(2x-1)(4x-1)$

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

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

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