How do you solve $x^{2} - \frac{1}{4} x - \frac{3}{64} = 64$ $x^2- 1/4x - 3/64=64$ by completing the square?

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How do you solve $x^{2} - \frac{1}{4} x - \frac{3}{64} = 64$ $x^2- 1/4x - 3/64=64$ by completing the square?

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Answer 1 · 2017-10-21T13:09:42

$x = \frac{1 \pm 10 \sqrt{41}}{8}$ $x=(1+-10sqrt(41))/8$ .

Explanation:

To complete the square, you can apply the formula $x^{2} + a x = {(x + \frac{1}{2} a)}^{2} - {(\frac{1}{2} a)}^{2}$ $x^2+ax=(x+1/2a)^2-(1/2a)^2$ .

You have to substitute $a = \frac{- \frac{1}{4}}{2} = - \frac{1}{8}$ $a=(-1/4)/2=-1/8$ to solve this problem.

$x^{2} - \frac{1}{4} x - \frac{3}{64} = 64$ $x^2-1/4x-3/64=64$
$x^{2} - \frac{1}{4} x = 64 + \frac{3}{64}$ $x^2-1/4x=64+3/64$
${(x - \frac{1}{8})}^{2} - {(- \frac{1}{8})}^{2} = 64 + \frac{3}{64}$ $(x-1/8)^2-(-1/8)^2=64+3/64$
${(x - \frac{1}{8})}^{2} = 64 + \frac{3}{64} + {(- \frac{1}{8})}^{2}$ $(x-1/8)^2=64+3/64+(-1/8)^2$
${(x - \frac{1}{8})}^{2} = 64 + \frac{4}{64} = \frac{1025}{16}$ $(x-1/8)^2=64+4/64=1025/16$

Notice that $1025 = 5^{2} \cdot 41$ $1025=5^2*41$ .
The root of the equation is
$x - \frac{1}{8} = \pm \sqrt{\frac{1025}{16}}$ $x-1/8=+-sqrt(1025/16)$
$x - \frac{1}{8} = \pm \frac{5 \sqrt{41}}{4}$ $x-1/8=+-(5sqrt(41))/4$
$x = \frac{1 \pm 10 \sqrt{41}}{8}$ $x=(1+-10sqrt(41))/8$ .

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How do you solve $x^{2} - \frac{1}{4} x - \frac{3}{64} = 64$ $x^2- 1/4x - 3/64=64$ by completing the square?

1 Answer

Explanation:

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How do you solve $x^{2} - \frac{1}{4} x - \frac{3}{64} = 64$ $x^2- 1/4x - 3/64=64$ by completing the square?