How do you solve $x^{2} - 2 x = 5$ $x^2-2x=5$ by completing the square?

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How do you solve $x^{2} - 2 x = 5$ $x^2-2x=5$ by completing the square?

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Answer 1 · 2016-07-11T04:11:59

$x = 1 \pm \sqrt{6}$ $x = 1 ± sqrt(6)$

Explanation:

Take the coefficient on the $x$ $x$ -term, namely $- 2$ $-2$ , divide by $2$ $2$ , and square the result, giving you

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

Thus, we can now replace every $?$ $?$ mark in the expression

$x^{2} - 2 x + ? = 5 + ?$ $x^2-2x + ? = 5 + ?$

with the number $1$ $1$ , giving us

$x^{2} - 2 x + 1 = 6$ $x^2-2x+1=6$

We'd like two numbers whose product is $1$ $1$ and when added together gives us the result of $- 2$ $-2$ (the number on the $x$ $x$ -term).

Since $(- 1) \cdot (- 1) = 1$ $(-1) * (-1) = 1$ and $(- 1) + (- 1) = - 2$ $(-1) + (-1) = -2$ , we now have our factors and can rewrite our expression in the following way:

$(x - 1) (x - 1) = 6$ $(x-1)(x-1) = 6$

or

${(x - 1)}^{2} = 6$ $(x-1)^2 = 6$

Taking the square root of both sides of the equation yields

$\sqrt{{(x - 1)}^{2}} = \pm \sqrt{6}$ $sqrt((x-1)^2) = ± sqrt(6)$

$(x - 1) = \pm \sqrt{6}$ $(x-1) = ± sqrt(6)$

Adding $1$ $1$ to both sides gives us our final result of

$x = 1 \pm \sqrt{6}$ $x = 1 ± sqrt(6)$

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How do you solve $x^{2} - 2 x = 5$ $x^2-2x=5$ by completing the square?

1 Answer

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