Question 118925
{{{x^2+8x+13=0}}}
First, get all the x terms on one side of the equation:
{{{x^2+8x=-13}}}
Divide the coefficient of x by 2 and square the result. Add this result to both sides of the equation:
{{{x^2+8x+16=-13+16}}}
Simplify right side of the equation:
{{{x^2+8x+16=3}}}
Express the right side as a perfect square:
{{{(x+4)^2=3}}}
Remember the goal is to solve for x. Take the square root of both sides of the equation:
(x+4)=+-sqrt(3)
Since this is a second order equation, x can assume 2 values. Take the case of the {{{+sqrt(3)}}} first and solve for x:
{{{x+4=sqrt(3)}}}
{{{highlight(x=sqrt(3)-4)}}}
Now, take the case of {{{-sqrt(3)}}}:
{{{x+4=-sqrt(3)}}}
{{{highlight(x=-sqrt(3)-4)}}}
Plug each of these results into the original equation to check and see if they are right.
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good Luck,
tutor_paul@yahoo.com