A rectangle's length is 5 m more than twice its width. Its area is 88 square metres. Find its perimeter.

Step-by-step solution

1. 1

   Name the width and build the length from it. Let the width be $x$ metres. "Five more than twice the width" is $2x+5$, so the length is $2x+5$ metres. Because the area condition multiplies the two sides, the resulting equation will be quadratic rather than linear.

2. 2

   Write the area equation.

   $x(2x+5)=88\quad\Longrightarrow\quad 2x^{2}+5x-88=0$

3. 3

   Solve the quadratic. The discriminant is

   $5^{2}-4(2)(-88)=25+704=729=27^{2}$

   a perfect square, so

   $x=\frac{-5\pm 27}{4}\quad\Longrightarrow\quad x=\frac{22}{4}=5.5\ \text{ or }\ x=-8$

4. 4

   Discard the impossible root. A width cannot be negative, so $x=-8$ is rejected on physical grounds and the width is $5.5$ m. The length is then

   $2(5.5)+5=16\ \text{m}$

5. 5

   Compute the perimeter.

   $P=2(\text{length}+\text{width})=2(16+5.5)=2(21.5)=43\ \text{m}$

   $\boxed{43\ \text{m}}$

6. 6

   Check both given conditions. Area: $16\times 5.5=88\ \text{m}^2$ ✓. Length rule: $2(5.5)+5=16$ ✓. Note the answer is a perimeter (metres), while the given datum was an area (square metres) — a unit change worth flagging in the final line.