How do you complete the square for $8 x^{2} - 160 x + 560$ $8x^2 -160x + 560$ ?

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Answer 1 · 2015-07-07T02:37:32

Solve $y = 8 x^{2} - 160 + 560 = 0$ $y = 8x^2 - 160 + 560 = 0$

$y = 8 (x^{2} - 20 x + 70) = 0$ $y = 8(x^2 - 20x + 70) = 0$

$(x^{2} - 20 x + 100) = 100 - 70 = 30$ $(x^2 - 20x + 100) = 100 - 70 = 30$

${(x - 10)}^{2} = 30$ $(x - 10)^2 = 30$ -> $(x - 10) = \pm \sqrt{30}$ $(x - 10) = +- sqrt30$

$x = 10 \pm \sqrt{30}$ $x = 10 +- sqrt30$

How do you complete the square for 8x^2 -160x + 5608x2−160x+5608x^2 -160x + 560?