How do you solve $- x^{2} - 8 x + 5 = 0$ $-x^2 - 8x + 5 = 0$ by completing the square?

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Answer 1 · 2016-06-27T20:47:38

$x = 0.583 or x = - 8.583$ $x =0.583 " or "x = -8.583$

The negative $x^{2}$ $x^2$ term is not comfortable. Divide by -1.

$x^{2} + 8 x - 5 = 0$ $x^2 +8x -5 =0$

Move the constant to the other side: $x^{2} + 8 x = 5$ $" "x^2 +8x = 5$

Add on the square of half the co-efficient of x to both sides to make the square of a binomial.

$x^{2} + 8 x + 16 = 5 + 16 {(\frac{8}{2})}^{2}$ $x^2 + 8x + color(red)16 = 5 + color(red)16 " "color(red)((8/2)^2$

${(x + 4)}^{2} = 21 square root both sides$ $(x+4)^2 = 21 " square root both sides"$

$x + 4 = \pm \sqrt{21}$ $x+4 = +-sqrt21$

$x = + \sqrt{21} - 4 or x = - \sqrt{21} - 4$ $x = +sqrt21 -4" or " x = -sqrt21 -4$

$x = 0.583 or x = - 8.583$ $x =0.583 " or "x = -8.583$

How do you solve −x2−8x+5=0-x^2 - 8x + 5 = 0 by completing the square?