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

Answer 1

The constant (#2#) is subtracted from both sides to get #x^2-4x=-2#.

#a=-2# and the term required to finish the square is #a^2 =4# if the first two terms of a square of the form #(x+a)^2# are #x^2 +4x#.

#x^2-4x+4 = 4-2# adds 4 to both sides of the equation.

Square up the left side to #(x-2)^2 = 2#.

Square both sides' roots to get #x-2 = +-sqrt(2)#.

Providing the answers: #x = 2+-sqrt(2)#

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Answer 2

Genesis Allen

To solve the equation (x^2 - 4x + 2 = 0) by completing the square, follow these steps:

Move the constant term to the other side of the equation: (x^2 - 4x = -2)
To complete the square, take half of the coefficient of (x) (which is (-4/2 = -2)), square it (((-2)^2 = 4)), and add it to both sides of the equation: (x^2 - 4x + 4 = -2 + 4)
Rewrite the left side as a perfect square: ((x - 2)^2 = 2)
Solve for (x): (x - 2 = \pm \sqrt{2})
Add 2 to both sides: (x = 2 \pm \sqrt{2})

So, the solutions are (x = 2 + \sqrt{2}) and (x = 2 - \sqrt{2}).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.