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

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Answer 1 · 2016-07-11T04:11:59

$x = 1 ± sqrt(6)$

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

$(-2/2)^(2) = (-1)^2 = 1$

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

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

with the number $1$ , giving us

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

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

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

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

or

$(x-1)^2 = 6$

Taking the square root of both sides of the equation yields

$sqrt((x-1)^2) = ± sqrt(6)$

$(x-1) = ± sqrt(6)$

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

$x = 1 ± sqrt(6)$

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