sloving quadratic equations by factoring

Question

Find the dimensions of a rectangle that is longer than it is wide whose perimeter is 22ft, and whose area is 18ft2

Answer 1

This is a problem which defines two different conditions that must be satisfied at the same time, from which we can write two equations.

Condition #1: The perimeter must be 22ft
Equation #1:  L + L + W + W = 22
                    2L + 2W = 22
                    2(L + W) = 22
                    L + W = 11

Condition #2: The area must be 18 ft²
Equation #2: L * W = 18

So we must solve the two equations simultaneously.  We can do this by substitution (there are other methods as well) which we do by solving one of the equations for one variable, and we substitute the answer into the other equation:

Step 1: Solve the equation for L
            L = 11 - W

Step 2: Plug into the other equation:
           (11 - W) * W = 18

Step 3: Solve for W
           11W - W² = 18
           W² - 11W + 18 = 0
           (W - 9)(W-2) = 0
           W=9   or   W=2

Step 4: Solve for the other variable using either of the two equations:
           If W=9 then L=2
     or   If W=2 then L=9

Since your problem statement requires the rectangle to be longer than it is wide, we choose the answer that L=9 and W=2.

Answer 2

This type of problem was solved frequently by ancients. It's the system of two equations:

L+W=S and LW=A (S = semiperimeter, A = area)

The solution is that L and W are roots of t² - St + A = 0

This is because because (t - L)(t - W) = 0 expands to t² - (L+W)t + LW = 0.

In this case, t² - 11t + 18 = 0 which factors as (t - 2)(t - 9) = 0 so t = 2 or t = 9.

Thus the rectangle is 9 feet long and 2 feet wide.

Answer 3

Mental math approach:

L+W = half of the perimeter.

LW = area

So, L+W = 11, L*W = 18

Factor 18 into two factors such that the sum of the two is 11.

18 = 2*9, and 2+9 = 11.

Therefore, the answer is width = 2 ft, and length = 9 ft.