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

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

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Answer 1 · 2017-06-28T12:28:40

$x = - 4 \pm \sqrt{14}$ $x=-4+-sqrt14$

Explanation:

$express as x^{2} + 8 x = - 2$ $"express as " x^2+8x=-2$

$to complete the square$ $"to "color(blue)"complete the square"$

add ${(\frac{1}{2} coefficient of x-term)}^{2} to both sides$ $(1/2"coefficient of x-term")^2" to both sides"$

$that is add {(\frac{8}{2})}^{2} = 16 to both sides$ $"that is add " (8/2)^2=16" to both sides"$

$\Rightarrow x^{2} + 8 x + 16 = - 2 + 16$ $rArrx^2+8xcolor(red)(+16)=-2color(red)(+16)$

$\Rightarrow {(x + 4)}^{2} = 14$ $rArr(x+4)^2=14$

$take the square root of both sides$ $color(blue)"take the square root of both sides"$

$\sqrt{{(x + 4)}^{2}} = \pm \sqrt{14} \leftarrow note plus or minus$ $sqrt((x+4)^2)=+-sqrt14larr" note plus or minus"$

$\Rightarrow x + 4 = \pm \sqrt{14}$ $rArrx+4=+-sqrt14$

$subtract 4 from both sides$ $"subtract 4 from both sides"$

$xcancel(+4)cancel(-4)=+-sqrt14-4$

$rArrx=-4+-sqrt14$

Answer 2 · 2017-06-28T12:30:01

Move +2 to the right side of the equation.
$x^2 + 8x = -2$

Then halve the coefficient of x.
$x^2 + (8/2)x = -2$

Then square that same coefficient.
$x^2 + (8/2)^2x = -2$

Since $(8/2)^2$ = $16$ we can put that number into the equation.
So, $x^2 + 8x + 16 = -2$

When you find the number to complete the square you must add it to both sides of the equation.
So, $x^2 + 8x + 16 = -2 +16$
= $x^2 + 8x + 16 = 14$

Then factorise $x^2 + 8x + 16$ = $(x+4)(x+4)$

Therefore, the answer is: $(x+4)(x+4)=14$

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

2 Answers

Explanation:

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