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#