Short Answer

Determining Real-World Sign Dimensions from Quadratic Solutions

As an office facilities coordinator managing a building upgrade, you order a custom rectangular logo sign with an area of 30 square feet. The sign maker specifies that the length of the sign must be exactly one foot more than the width (ww). To verify the required space on the lobby wall, you solve the resulting quadratic equation w2+w30=0w^2 + w - 30 = 0 and find two algebraic solutions: w=5w = 5 and w=6w = -6.

Based on these results, state the final physical dimensions (width and length in feet) of the lobby sign, and briefly explain why one of the algebraic solutions was excluded.

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Updated 2026-05-31

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