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

Question

26785 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-08T07:36:13

Force a perfect square trinomial on the left side. Take the square root of both sides and solve for $x$ .

$x^2+10x-2=0$

Add $2$ to both sides of the equation.

$x^2+10x=2$

Force a perfect square trinomial on the left side of the equation.

Divide the coefficient of the $x$ term and square the result. Add it to both sides of the equation.

$(10/2)^2=5^2=25$

$x^2+10x+25=2+25$ =

$x^2+10x+25=27$

We now have a perfect square trinomial on the left side with the form $a^2+2ab+b^2=(a+b)^2$ .

$a=x,$ $b=5$

$(x+5)^2=27$

Take the square root of both sides.

$(x+5)=+-sqrt 27$ =

$x+5=+-sqrt 3sqrt 9$ =

$x+5=+-3sqrt 3$ =

Subtract $5$ from both sides.

$x=-5+-3sqrt3$

$x=-5+3sqrt 3$

$x=-5-3sqrt3$

$x=-5+3sqrt 3, -5-3sqrt3$

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