How do you solve $x^{2} - 10 x + 8 = 0$ $x^2-10x+8=0$ by completing the square?

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How do you solve $x^{2} - 10 x + 8 = 0$ $x^2-10x+8=0$ by completing the square?

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Answer 1 · 2015-04-29T17:34:09

Write it as:
$x^{2} - 10 x = - 8$ $x^2-10x=-8$ add and subtract $25$ $25$ ;
$x^{2} - 10 x + 25 - 25 = - 8$ $x^2-10x+25-25=-8$
${(x - 5)}^{2} - 25 = - 8$ $(x-5)^2-25=-8$
${(x - 5)}^{2} = 17$ $(x-5)^2=17$
$x - 5 = \pm \sqrt{17}$ $x-5=+-sqrt(17)$
$x_{1, 2} = 5 \pm \sqrt{17}$ $x_(1,2)=5+-sqrt(17)$

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How do you solve $x^{2} - 10 x + 8 = 0$ $x^2-10x+8=0$ by completing the square?

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How do you solve x2−10x+8=0x^2-10x+8=0 by completing the square?

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How do you solve $x^{2} - 10 x + 8 = 0$ $x^2-10x+8=0$ by completing the square?