How do you solve $x^{2} - 8 x + 3 = 0$ $x^2 - 8x + 3 = 0$ by completing the square?

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How do you solve $x^{2} - 8 x + 3 = 0$ $x^2 - 8x + 3 = 0$ by completing the square?

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Answer 1 · 2015-10-08T16:44:03

The solutions are $4 \pm \sqrt{13}$ $4 \pm sqrt(13)$ .

Explanation:

We must manipulate the expression into something of the form

$x^{2} + 2 a x + a^{2}$ $x^2 + 2ax+ a^2$

Since the term $2 a x$ $2ax$ equals $- 8 x$ $-8x$ , it means that $a = - 4$ $a=-4$ , and thus $a^{2} = 16$ $a^2=16$ . We need $13$ $13$ more units for the constant to reach that value, so we can add and subtract:

$x^{2} - 8 x + 3 + 13 - 13 = 0 \to x^{2} - 8 x + 16 = 13$ $x^2 -8x+ 3 + 13 - 13 = 0 -> x^2 -8x +16 =13$

But now we have that $x^{2} - 8 x + 16 = {(x - 4)}^{2}$ $x^2 -8x +16 = (x-4)^2$ , so the equation becomes

${(x - 4)}^{2} = 13$ $(x-4)^2 = 13$

Taking square roots of both sides:

$x - 4 = \pm \sqrt{13}$ $x-4 = \pm \sqrt(13)$

$x = 4 \pm \sqrt{13}$ $x=4 \pm \sqrt(13)$

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How do you solve $x^{2} - 8 x + 3 = 0$ $x^2 - 8x + 3 = 0$ by completing the square?

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Explanation:

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How do you solve x^2 - 8x + 3 = 0x2−8x+3=0x^2 - 8x + 3 = 0 by completing the square?

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How do you solve $x^{2} - 8 x + 3 = 0$ $x^2 - 8x + 3 = 0$ by completing the square?