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

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Answer 1 · 2018-04-17T20:18:37

$x = - \frac{5}{2} + \frac{\sqrt{17}}{2}$ $x=-5/2+sqrt17/2$ and $x = - \frac{5}{2} - \frac{\sqrt{17}}{2}$ $x=-5/2-sqrt17/2$

Half the middle term, add it to $x$ $x$ and square it. Keep the constant on the end

$x^{2} + 5 x \to {(x + \frac{5}{2})}^{2} + 2$ $x^2+5x -> (x+5/2)^2+2$

Take away the term previously halved, but squared.

${(x + \frac{5}{2})}^{2} - \frac{25}{4} + \frac{8}{4}$ $(x+5/2)^2-25/4+8/4$

Turn the extra constant to have the same denominator to make simplifying easier

Simplify:

${(x + \frac{5}{2})}^{2} - \frac{17}{4} = 0$ $(x+5/2)^2-17/4=0$

Solving:

Plus the constant, making it cancel out:

$(x+5/2)^2cancel(-17/4)=17/4$

Get rid of the square bracket by square rooting:

$x+5/2=pmsqrt(17/4)$ $larr$ Notice the $pm$ sign, we add this when square rooting.

Minus the constant:

$x+cancel(5/2)=-5/2pmsqrt(17/4)$

Simplify (If needed):

$sqrt4 -> 2$

$x=-5/2pmsqrt17/2$

Therefore our 2 answers are:

$x=-5/2+sqrt17/2$ and $x=-5/2-sqrt17/2$

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