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

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Answer 1 · 2015-06-11T22:00:11

$0 = x^{2} - 4 x + 2 = {(x - 2)}^{2} - 2$ $0 = x^2-4x+2 = (x-2)^2-2$ .

Hence $x - 2 = \pm \sqrt{2}$ $x-2 = +-sqrt(2)$ and $x = 2 \pm \sqrt{2}$ $x = 2+-sqrt(2)$

${(x - 2)}^{2} = x^{2} - 4 x + 4$ $(x-2)^2 = x^2-4x+4$

So $x^{2} - 4 x + 2 = {(x - 2)}^{2} - 4 + 2 = {(x - 2)}^{2} - 2$ $x^2 - 4x+2 = (x-2)^2 - 4 + 2 = (x-2)^2-2$

To solve ${(x - 2)}^{2} - 2 = 0$ $(x-2)^2-2 = 0$ , first add $2$ $2$ to both sides to get:

${(x - 2)}^{2} = 2$ $(x-2)^2 = 2$

Then $(x - 2) = \pm \sqrt{2}$ $(x-2) = +-sqrt(2)$

Add $2$ $2$ to both sides to get:

$x = 2 \pm \sqrt{2}$ $x = 2+-sqrt(2)$

In the general case:

$a x^{2} + b x + c$ $ax^2 + bx + c$

$= a {(x + \frac{b}{2 a})}^{2} + (c - \frac{b^{2}}{4 a})$ $=a(x+b/(2a))^2 + (c - b^2/(4a))$

From which we can deduce the quadratic formula:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b +- sqrt(b^2-4ac)) / (2a)$

How do you solve x2−4x+2=0x^2 - 4x +2 = 0 using completing the square?