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The length of a rectangle is 3 inches longer than it is wide. If the area is 130 square inches, what are the dimensions of the rectangle? The width, or shorter side is ◻ inches The length, or longer side is ◻ inches

The length of a rectangle is 33 inches longer than it is wide. If the area is 130130 square inches, what are the dimensions of the rectangle?\newlineThe width, or shorter side is □ \square inches\newlineThe length, or longer side is □ \square inches

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Full solution

Q. The length of a rectangle is 33 inches longer than it is wide. If the area is 130130 square inches, what are the dimensions of the rectangle?\newlineThe width, or shorter side is □ \square inches\newlineThe length, or longer side is □ \square inches
  1. Define Width and Length: Let's define the width of the rectangle as WW inches. Since the length is 33 inches longer, the length will be W+3W + 3 inches.
  2. Calculate Area Equation: The area of a rectangle is calculated by multiplying the length by the width. So, the equation for the area is W×(W+3)=130W \times (W + 3) = 130 square inches.
  3. Rearrange and Solve Quadratic Equation: Expanding the equation gives us W2+3W=130W^2 + 3W = 130. To solve for WW, we need to rearrange this into a standard quadratic form. Subtract 130130 from both sides to get W2+3W−130=0W^2 + 3W - 130 = 0.
  4. Apply Quadratic Formula: Now, we solve the quadratic equation W2+3W−130=0W^2 + 3W - 130 = 0 using the quadratic formula, W=−b±b2−4ac2aW = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. Here, a=1a = 1, b=3b = 3, and c=−130c = -130.
  5. Simplify and Find Potential Solutions: Plugging in the values, we get W=−3±32−4⋅1⋅(−130)2⋅1W = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-130)}}{2 \cdot 1}. Simplifying inside the square root: W=−3±9+5202W = \frac{-3 \pm \sqrt{9 + 520}}{2}.
  6. Select Valid Width: Further simplifying, W=−3±5292W = \frac{-3 \pm \sqrt{529}}{2}. Since 529=23\sqrt{529} = 23, we have W=−3±232W = \frac{-3 \pm 23}{2}.
  7. Calculate Length: This gives us two potential solutions for WW: W=(23−3)/2=10W = (23 - 3) / 2 = 10 and W=(−3−23)/2=−13W = (-3 - 23) / 2 = -13. Since a width can't be negative, we use W=10W = 10 inches.
  8. Calculate Length: This gives us two potential solutions for WW: W=(23−3)/2=10W = (23 - 3) / 2 = 10 and W=(−3−23)/2=−13W = (-3 - 23) / 2 = -13. Since a width can't be negative, we use W=10W = 10 inches. Now, substituting W=10W = 10 inches back into the length equation, the length L=W+3=10+3=13L = W + 3 = 10 + 3 = 13 inches.

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