Word problems (help please!)

Question

A square is altered so that one dimension is increased by 4, while the other dimension is decreased by 3. The area of the resulting rectangle is 60. Find the area of the original square.

Answer 1

Let x be the sides of a square.

Area = (x+4)(x-3) = 60

x²+x-12 = 60

x²+x-72 = 0

(x+9)(x-8) = 0 --> x = -9, 8 choose the positive one

Area of the original = x² = 64.

Answer 2

Notice that you have a square so since both the length and width of a square are the same lengths call both the length and the width x.

one dimension increased by 4: x+4 -> length

one dimension decreased by 3: x-3 -> width

Area of the resulting rectangle is 60: Area = length * width -> 60 = (x+4)(x-3)

Find the area of the original square: Find x²

Steps to solve:

(1) Solve for x using the formula for the area of the resulting triangle

60 = (x+4)(x-3) *distribute and set equal to 0

60 = x²+x-12

x²+x-72=0 *factor

(x-8)(x+9) = 0 *solve

x=8, x=-9 *only the positive one counts since lengths can't be negative

x=8

(2) Find x²

(8²) = 64

Answer: Area of square = 64

Answer 3

(s + 4)(s - 3) = 60

s² + s - 12 = 60

Can you solve for s and answer?