How do you solve #x^{2}+(x+1)^{2}=25#?

1 Answer
Mark
Mar 27, 2017

#x= -4, 3#

Explanation:

#x^2 + (x+1)^2 = 25#
#x^2 + (x+1)(x+1) = 25#
#x^2 + x^2+2x+1 = 25#
#2x^2 +2x-24 = 0# (Divide by a #2#)
#x^2 +x-12 = 0#
#(x+4)(x-3)=0#
#x+4= 0#
#x-3=0#
#x= -4, 3#

Plug in to make sure:
#x^2 + (x+1)^2 = 25#
#-4^2 + (-4+1)^2 = 25#
#16 + 9 = 25#
#25=25#

#3^2 + (3+1)^2 = 25#
#9 + 16 = 25#
#25=25#

Both are correct, so # x = -4 and 3#

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