1Cademy - Finding the Dimensions of a 150 Square Foot Rectangular Turf Area

Learn Before

How to Solve a Quadratic Equation Using the Quadratic Formula

Example

Finding the Dimensions of a 150 Square Foot Rectangular Turf Area

Apply the seven-step problem-solving strategy to a rectangle area problem where one dimension is expressed in terms of the other using both multiplication and subtraction, producing a quadratic equation whose discriminant is not a perfect square and whose solution must be approximated.

Problem: Mike wants to put $150$ square feet of artificial turf in his front yard. He wants a rectangular area with length one foot less than three times the width. Find the length and width, rounded to the nearest tenth of a foot.

Read: A rectangular turf area has $A = 150$ sq ft, and its length is one foot less than three times the width. Draw and label the rectangle with width $w$ and length $3w - 1$ .
Identify: The length and width of the rectangle.
Name: Let $w$ = the width. Then $3w - 1$ = the length.
Translate: Write the area formula and substitute:

$A = L \cdot W$

$150 = (3w - 1)(w)$

Solve: Distribute: $150 = 3w^2 - w$ . Rewrite in standard form: $3w^2 - w - 150 = 0$ . Identify coefficients: $a = 3$ , $b = -1$ , $c = -150$ . Substitute into the Quadratic Formula:

$w = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(-150)}}{2(3)} = \frac{1 \pm \sqrt{1 + 1800}}{6} = \frac{1 \pm \sqrt{1801}}{6}$

Since $\sqrt{1801}$ does not simplify, write the two solutions:

$w = \frac{1 + \sqrt{1801}}{6} \approx 7.2 \quad \text{or} \quad w = \frac{1 - \sqrt{1801}}{6} \approx -6.9$

Because $w$ represents a physical width, $w \approx -6.9$ is discarded. So $w \approx 7.2$ , and the length is $3(7.2) - 1 = 20.6$ .

Check: $7.2 \times 20.6 = 148.32$ , which is close to $150$ . The small discrepancy is due to rounding.
Answer: The width of the rectangle is approximately $7.2$ feet and the length is approximately $20.6$ feet.

Unlike the earlier rectangular garden example — where the quadratic factored neatly and produced integer solutions — this problem has a discriminant of $1801$ , which is not a perfect square. When the discriminant is not a perfect square, the solutions involve irrational numbers that must be approximated with a calculator. This demonstrates that real-world area problems do not always produce "clean" answers, and the Quadratic Formula handles these cases where factoring cannot.

Updated 2026-04-21

Contributors are:

Who are from:

References

OpenStax Elementary Algebra

Learn Before

Related

Learn After