If #(a-b)^2=40# and #ab=8#, what is #a^2+b^2#?

2 Answers
Feb 23, 2016

56

Explanation:

distribute #(a - b )^2" using FOIL (or any method you use) "#

#(a - b )^2 = a^2 - 2ab + b^2 = 40 #

( substitute ab = 8 )

hence #a^2 - 2(8) + b^2 = 40 → a^2 - 16 + b^2 = 40#

#rArr a^2 - 16 + b^2 = 40 rArr a^2 + b^2 = 40 + 16 = 56 #

Feb 23, 2016

#56#

Explanation:

#color(blue)((a-b)^2=40,ab=8,a^2+b^2=?#

Remember the formula #color(orange)((a-b)^2=a^2-2ab+b^2#

So replace #(a-b)^2# by #a^2-2ab+b^2# in the equation

#rarra^2-2ab+b^2=40#

We know that #ab=8#

So,replace #ab# by #8# in the equation

#rarra^2-2(8)+b^2=40#

#rarra^2-16+b^2=40#

Add 16 both sides

We get,

#rarra^2+b^2=40+16#

#color(green)(rArra^2+b^2=56#