QUESTION IMAGE
Question
- marias garden has a rectangular shape with a length that is 3 feet longer than twice the width. if the area of the garden is 65 square feet, then what are the dimensions of the garden?
Step1: Define variables for width
Let width = $w$
Step2: Express length in terms of $w$
Length = $2w + 3$
Step3: Set up area equation
Area = length $\times$ width, so:
$$w(2w + 3) = 65$$
Step4: Expand and rearrange to quadratic
$$2w^2 + 3w - 65 = 0$$
Step5: Solve quadratic equation
Use quadratic formula $w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=2$, $b=3$, $c=-65$:
$$w = \frac{-3 \pm \sqrt{3^2 - 4(2)(-65)}}{2(2)}$$
$$w = \frac{-3 \pm \sqrt{9 + 520}}{4}$$
$$w = \frac{-3 \pm \sqrt{529}}{4}$$
$$w = \frac{-3 \pm 23}{4}$$
Take positive root (width can't be negative):
$$w = \frac{-3 + 23}{4} = \frac{20}{4} = 5$$
Step6: Calculate length
Length = $2(5) + 3 = 13$
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The width of the garden is 5 feet, and the length is 13 feet.