How do you solve the equation $x^2-10x+25=49$ by completing the square?

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Answer 1 · 2016-12-13T22:41:37

$x=12" "$ or $" "x = -2$

The difference of squares identity can be written:

$a^2-b^2 = (a-b)(a+b)$

Use this with $a=(x-5)$ and $b=7$ later.

Given:

$x^2-10x+25=49$

Both sides of this equation are already perfect squares:

$(x-5)^2 = x^2-10x+25 = 49 = 7^2$

Subtract $7^2$ from both ends to get:

$0 = (x-5)^2-7^2 = ((x-5)-7)((x-5)+7) = (x-12)(x+2)$

Hence:

$x=12" "$ or $" "x = -2$

How do you solve the equation x^2-10x+25=49x^2-10x+25=49 by completing the square?