FOIL is just distributive property of multiplication
(a+b)(c+d) = ac + ad + bc + bd
proof with distributive property:
(a+b)(c+d) = a(c+d) + b(c+d)
= ac + ad + bc + bd
CAN SOMEONE PLEASE HELP ME WITH MY SQUARE ROOT AND FOILING PLEASE?
5 answers
[�ãv(2x+2) - �ãv(x-3)][�ãv(2x+2) - �ãv(x-3)] = 4
I do not understand your special symbols but perhaps they are radical signs?
If so then
[ sqrt(2x+2) - sqrt (x-3) ] [sqrt(2x+2) -sqrt (x-3) ] = 4
sqrt (a) sqrt (a) = a
so
(2x+2) - 2 sqrt (2x+2) sqrt(x-3) + (x-3) = 4
or
3 x -2 sqrt (2x+2) sqrt(x-3) = 5
or
(3 x -5)/2 = sqrt (2x^2 -4 x - 6)
square both sides
(9 x^2 -30 x +25)/4 = 2 x^2 -4 x - 6
or
9 x^2 - 30 x + 25 = 8 x^2 - 16 x - 24
or
x^2 -14 x + 49 = 0
(x-7)(x-7) = 0
x = 7
If so then
[ sqrt(2x+2) - sqrt (x-3) ] [sqrt(2x+2) -sqrt (x-3) ] = 4
sqrt (a) sqrt (a) = a
so
(2x+2) - 2 sqrt (2x+2) sqrt(x-3) + (x-3) = 4
or
3 x -2 sqrt (2x+2) sqrt(x-3) = 5
or
(3 x -5)/2 = sqrt (2x^2 -4 x - 6)
square both sides
(9 x^2 -30 x +25)/4 = 2 x^2 -4 x - 6
or
9 x^2 - 30 x + 25 = 8 x^2 - 16 x - 24
or
x^2 -14 x + 49 = 0
(x-7)(x-7) = 0
x = 7
Thank you.
x- -1 when x=3