Question 1151211
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Square A has side length x+3
Square B has side length x
Square A has a side length 3 units larger compared to the side length of square B.
Negative side lengths are not possible, so x > 0.


Compute the area of square A
area of square A = (side length of square A)^2
area of square A = (x+3)^2
area of square A = (x+3)(x+3)
area of square A = x(x+3)+3(x+3)
area of square A = x^2+3x+3x+9
area of square A = x^2+6x+9
You could use the FOIL rule to expand out (x+3)^2, but I think the distributive property is more versatile. 


Compute the area of square B
area of square B = (side length of square B)^2
area of square B = x^2


Add up the two areas
total area = (area of square A)+(area of square B)
total area = (x^2+6x+9)+(x^2)
total area = 2x^2+6x+9


Set this total area equal to 45 and solve for x
2x^2+6x+9 = 45
2x^2+6x+9-45 = 0
2x^2+6x-36 = 0
2(x^2+3x-18) = 0
x^2+3x-18 = 0/2
x^2+3x-18 = 0
(x+6)(x-3) = 0
x+6 = 0 or x-3 = 0
x = -6 or x = 3


Since x > 0, this means we only consider x = 3 as the solution.
If x = 3, then x+3 = 3+3 = 6.


Square A has a side length of 6 inches and an area of 36 square inches.
Square B has a side length of 3 inches and an area of 9 square inches.
The total area is 36+9 = 45 square inches.
The answer checks out.
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