How do you complete the square to solve $x^{2} + 5 x + 6 = 0$ $x^2 + 5x + 6 = 0$ ?

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How do you complete the square to solve $x^{2} + 5 x + 6 = 0$ $x^2 + 5x + 6 = 0$ ?

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Answer 1 · 2015-06-16T23:41:19

You can complete the square by first getting the form $x^{2} + k x = h$ $x^2 + kx = h$ .

$x^{2} + 5 x = - 6$ $x^2 + 5x = -6$

Then just add and subtract a certain value that is equal to ${(\frac{k}{2})}^{2}$ $(k/2)^2$ . Just remember that the function is $x^{2} + k x$ $x^2 + kx$ , so $k$ $k$ may be negative, but the added ${(\frac{k}{2})}^{2}$ $(k/2)^2$ will always be positive.

$x^{2} + 5 x + {(\frac{5}{2})}^{2} = {(\frac{5}{2})}^{2} - 6$ $x^2 + 5x + (5/2)^2 = (5/2)^2 - 6$

${(x + \frac{5}{2})}^{2} = \frac{25}{4} - \frac{24}{4} = \frac{1}{4}$ $(x+5/2)^2 = 25/4 - 24/4 = 1/4$

${(x + \frac{5}{2})}^{2} - \frac{1}{4} = 0$ $(x+5/2)^2 - 1/4 = 0$

graph{(x+5/2)^2 - 1/4 [-10, 10, -5, 5]}

If you wanted to solve this:

${(x + \frac{5}{2})}^{2} = \frac{1}{4}$ $(x+5/2)^2 = 1/4$

$x + \frac{5}{2} = \pm \sqrt{\frac{1}{4}}$ $x+5/2 = pmsqrt(1/4)$

Thus:
$x + \frac{5}{2} = \pm \sqrt{\frac{1}{4}}$ $x+5/2 = pmsqrt(1/4)$

$x = \pm (\sqrt{\frac{1}{4}}) - \frac{5}{2}$ $x = pm(sqrt(1/4)) - 5/2$

$x = \pm (\frac{1}{2}) - \frac{5}{2}$ $x = pm(1/2) - 5/2$

$x = \frac{1}{2} - \frac{5}{2} = - 2$ $x = 1/2 - 5/2 = -2$

$x = - \frac{1}{2} - \frac{5}{2} = - 3$ $x = -1/2 - 5/2 = -3$

${(- 2)}^{2} + 5 (- 2) + 6 = 4 - 10 + 6 = 0$ $(-2)^2 + 5(-2) + 6 = 4 - 10 + 6 = 0$
${(- 3)}^{2} + 5 (- 3) + 6 = 9 - 15 + 6 = 0$ $(-3)^2 + 5(-3) + 6 = 9 - 15 + 6 = 0$

You could just factor, though...

$(x + 2) (x + 3) = x^{2} + 2 x + 3 x + 6 = x^{2} + 5 x + 6$ $(x+2)(x+3) = x^2 + 2x + 3x + 6 = x^2 + 5x + 6$

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How do you complete the square to solve $x^{2} + 5 x + 6 = 0$ $x^2 + 5x + 6 = 0$ ?

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