How do you solve #x^2 = 14x – 49#?

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Answer 1 · 2015-07-08T08:25:45

The answer is #x = 7#.

#x^2=14x-49#

Subtract #14x# from both sides of the equation.

#x^2-14x=-49#

Add #49# to both sides of the equation.

#x^2-14x+49=0#

#x^2-14x+49# is a perfect square trinomial in the form #a^2+2ab+b^2=(a+b)^2#, where #a=x and b=-7#.

#x^2-14x+49=0# =

#(x-7)^2=0#

Take the square root of both sides.

#(x-7)=sqrt 0# =

#x-7=0# =

#x=7#