How do you solve #2(x-5)^2 + 13 = 31#?

1 Answer
EZ as pi
Sep 5, 2016

#x = 2 or x = 8#

Explanation:

Although this is a quadratic equation, it is a special case because there is no term in #x#.

  • Move the number terms to the right side.

#2(x-5)^2 = 31-13 = 18#

  • Divide by 2 to isolate the bracket.

#(x-5)^2 = 9#

  • Find the square root of both sides and solve for x.

#x-5 = +-sqrt9 = +-3#

#x = +3+5 = 8#

or

#x= -3+5 = 2#

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