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

Question

Hazel Anglin · Accepted Answer

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# Make sure the left side of the equation has a perfect square trinomial. 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 each side's square root. #(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#

Kennedy Adams · Answer

undefined To solve $x^2 + 10x - 2 = 0$ by completing the square, follow these steps:

1. Move the constant term to the other side of the equation:
   $x^2 + 10x = 2$

2. Take half of the coefficient of $x$, square it, and add it to both sides of the equation:
   $x^2 + 10x + \left(\frac{10}{2}\right)^2 = 2 + \left(\frac{10}{2}\right)^2$

3. Simplify both sides:
   $x^2 + 10x + 25 = 2 + 25$

4. Factor the left side and simplify the right side:
   $(x + 5)^2 = 27$

5. Take the square root of both sides:
   $x + 5 = \pm \sqrt{27}$

6. Subtract 5 from both sides:
   $x = -5 \pm \sqrt{27}$

So the solutions are $x = -5 + \sqrt{27}$ and $x = -5 - \sqrt{27}$.