How do you solve $x^{2} - 2 x = 1$ $x^2 - 2x = 1$ using completing the square?

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How do you solve $x^{2} - 2 x = 1$ $x^2 - 2x = 1$ using completing the square?

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Answer 1 · 2015-06-19T19:49:28

I found:
$x_{1} = 1 + \sqrt{2}$ $x_1=1+sqrt(2)$
$x_{2} = 1 - \sqrt{2}$ $x_2=1-sqrt(2)$

Explanation:

Add and subtract $1$ $1$ to get:
$x^{2} - 2 x + 1 - 1 = 1$ $x^2-2xcolor(red)(+1-1)=1$
so:
$x^{2} - 2 x + 1 = 1 + 1$ $x^2-2xcolor(red)(+1)=1color(red)(+1)$
${(x - 1)}^{2} = 2$ $(x-1)^2=2$
$x - 1 = \pm \sqrt{2}$ $x-1=+-sqrt(2)$
$x = 1 \pm \sqrt{2}$ $x=1+-sqrt(2)$
$x_{1} = 1 + \sqrt{2}$ $x_1=1+sqrt(2)$
$x_{2} = 1 - \sqrt{2}$ $x_2=1-sqrt(2)$

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How do you solve $x^{2} - 2 x = 1$ $x^2 - 2x = 1$ using completing the square?

1 Answer

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How do you solve $x^{2} - 2 x = 1$ $x^2 - 2x = 1$ using completing the square?