How do you solve x^2 + 10x + 5 = 0 x2+10x+5=0 by completing the square?

1 Answer
Alan P.
May 13, 2016

x=-5+-2sqrt(5)x=−5±2√5
(see below for completing the squares method of solution)

Explanation:

Given:
color(white)("XXX")x^2+10x+5=0XXXx2+10x+5=0

Move the constant to the right side as
color(white)("XXX")color(blue)(x^2+10x)=-5XXXx2+10x=−5

We know that (x+a)^2=color(red)(x^2+2ax+a^2)(x+a)2=x2+2ax+a2
So if the first two terms of a squared binomial are
color(white)("XXX")color(red)(x^2+2ax) = color(blue)(x^2+10x)XXXx2+2ax=x2+10x
then
color(white)("XXX")color(red)(a)=color(blue)(5)XXXa=5
and we will need to add
color(white)("XXX")color(red)(a^2)=color(blue)(25)XXXa2=25 (to both sides) to complete the square:

color(white)("XXX")x^2+10x+25 = -5+25XXXx2+10x+25=−5+25

Writing as a squared binomial and simplifying the right side:
color(white)("XXX")(x+5)^2=20XXX(x+5)2=20

Taking the square root of both sides:
color(white)("XXX")x+5=+-2sqrt(5)XXXx+5=±2√5

Subtracting 55 from both sides
color(white)("XXX")x=-5+-2sqrt(5)XXXx=−5±2√5

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