If #a-b=10# and #2*(a+b)^2 = 64# then how do you find #a+b# ?

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Answer 1 · 2017-01-23T00:45:17

#a+b = +-4sqrt(2)#

Given that:

#2*(a+b)^2 = 64#

Divide both sides by #2# to find:

#(a+b)^2 = 32#

Hence:

#a+b = +-sqrt(32) = +-sqrt(4^2*2) = +-4sqrt(2)#

#color(white)()#
Bonus

Given:

#a-b=10#

#a+b=+-4sqrt(2)#

Adding these two equations, we find:

#2a = 10+-4sqrt(2)#

So #a = 5+-2sqrt(2)#

Then #b = a-10 = -5+-2sqrt(2)#

That is, the two possible solutions are:

#(a, b) = (5+2sqrt(2), -5+2sqrt(2))#

#(a, b) = (5-2sqrt(2), -5-2sqrt(2))#