How do you solve #x^2 - 4x +2 = 0# using completing the square?

Question

How do you solve #x^2 - 4x  +2 = 0# using completing the square?

Savannah Anderson · Accepted Answer

#0 = x^2-4x+2 = (x-2)^2-2#.

Hence #x-2 = +-sqrt(2)# and #x = 2+-sqrt(2)# #(x-2)^2 = x^2-4x+4# So #x^2 - 4x+2 = (x-2)^2 - 4 + 2 = (x-2)^2-2# To solve #(x-2)^2-2 = 0#, first add #2# to both sides to get: #(x-2)^2 = 2# Then #(x-2) = +-sqrt(2)# Add #2# to both sides to get: #x = 2+-sqrt(2)# In most cases: #ax^2 + bx + c# #=a(x+b/(2a))^2 + (c - b^2/(4a))# from which the quadratic formula can be derived: #x = (-b +- sqrt(b^2-4ac)) / (2a)#

Julia Adams · Answer

undefined To solve the equation $x^2 - 4x + 2 = 0$ using completing the square, follow these steps:

1. Move the constant term to the other side of the equation:
   $x^2 - 4x = -2$

2. Take half of the coefficient of $x$ (which is -4), square it, and add it to both sides of the equation:
   $x^2 - 4x + (-4/2)^2 = -2 + (-4/2)^2$
   $x^2 - 4x + 4 = -2 + 4$

3. Rewrite the left side as a perfect square trinomial:
   $(x - 2)^2 = 2$

4. Take the square root of both sides and solve for $x$:
   $x - 2 = \pm \sqrt{2}$
   $x = 2 \pm \sqrt{2}$

Therefore, the solutions to the equation are $x = 2 + \sqrt{2}$ and $x = 2 - \sqrt{2}$.