Quadratic equations

Question

A square is altered so that one domension is increased by 5 units, while the other dimension is decreased by 3 units. The area of the resulting rectangle is 48 sq units. Find the area of the original square.

Answer 1

The square will have the same length and width. In this case, let's say the squares side is of length x. One side was increased by 5 and one was decreased by 3 so the new rectangle has the following:

length = x+5

width = x-3

The area is 48 and can be obtained by multiplying length*width so you get the below equation:

(x+5)(x-3) = 48. using foil and putting like terms alike you get:

x²+2x-63 = 0. Now you can find the roots of the quadratic equation using the below formula:

So substituting the values you get x₁ = -9 and x₂= 7. Since a length cannot be of negative value, you can disregard x₁.

Now we know our side of the square is equal to x₂= 7, so the area of the original square would be 7*7 = 49.

Answer 2

Since this is a square to begin with, both sides can be represented by x. After the sides are altered, one side will be x + 5 and the other will be x - 3. So the area of the rectangle will be represented by (x + 5)(x - 3).

We know the area is 48 sq units so (x + 5)(x - 3) = 48. When we multiply the two binomials together, we get

x² + 2x -15 = 48. Now we have a quadratic equation to solve so the equation needs to equal 0. That means we subtract 48 from both sides.

x² + 2x - 63 = 0. This quadratic will factor to (x + 9)(x - 7) = 0.

Setting each factor equal to 0 gets us x = -9 and 7.We can't have a negative length so x = 7.

This means the area of the original square equals 7² which is 49.