How do you solve $x^{2} - 2 x + 2$ $x^2-2x+2$ by completing the square?

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How do you solve $x^{2} - 2 x + 2$ $x^2-2x+2$ by completing the square?

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Answer 1 · 2015-06-12T02:51:57

$x = 1 + i$ $x=1+i$

$x = 1 - i$ $x=1-i$

Explanation:

Make a perfect square trinomial by completing the square. A perfect square trinomial has the form of $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ $a^2+2ab+b^2=(a+b)^2$ or $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ $a^2-2ab+b^2=(a-b)^2$ .

$x^{2} - 2 x + 2 = 0$ $x^2-2x+2=0$

Subtract $2$ $2$ from both sides of the equation.

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

Divide the coefficient of the $x$ $x$ term by $2$ $2$ and square the result. Add it to both sides of the equation.

$\frac{- 2}{2} = - 1$ $(-2)/2=-1$ ; $- 1^{2} = 1$ $-1^2=1$

$x^{2} - 2 x + 1 = - 2 + 1$ $x^2-2x+1=-2+1$ =

$x^{2} - 2 x + 1 = - 1$ $x^2-2x+1=-1$

There is now a perfect square trinomial on the left side of the equation.

Factor the trinomial.

${(x - 1)}^{2} = - 1$ $(x-1)^2=-1$

Take the square root of both sides and solve for $x$ $x$ .

$x - 1 = \pm \sqrt{- 1}$ $x-1=+-sqrt(-1)$

$x - 1 = \pm i$ $x-1=+-i$

$x = 1 \pm i$ $x=1+-i$

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How do you solve $x^{2} - 2 x + 2$ $x^2-2x+2$ by completing the square?

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