Question

dimensions (length and width) of new market need to be? prove your answer in more than one way. step 1: write the inequality in standard form. (make sure one side is zero.) step 2: factor the quadratic expression. (break it down into two binomials.) step 3: find the roots. (set each factor equal to zero to find the boundary points.) step 4: test the intervals (use the roots to divide the number line into intervals and test one value from each.) step 5: write the solution set. solution: ____ pts w² + w - 132 ≥ 0 w² + w - 132 = (w + 12)(w - 11) w = -12, w = 11 (graph with test points) width = w ≥ 11 l = w + 1 ≥ 12

Explanation:

Step1: Correct Standard Form

The original inequality in Step 1 has a typo. The quadratic should be \( w^2 + w - 132 \geq 0 \) (assuming it's a quadratic in \( w \), likely a typo with \( w^2 \) instead of \( w + w \)).

Step2: Correct Factoring

For \( w^2 + w - 132 \), find two numbers that multiply to \(-132\) and add to \(1\). Those numbers are \(12\) and \(-11\). So factoring gives \( (w + 12)(w - 11) \), which matches the given (after correcting the quadratic term), so this step is correct if the quadratic was meant to be \( w^2 + w - 132 \).

Step3: Correct Roots

Setting \( (w + 12)(w - 11) = 0 \), we get \( w = -12 \) or \( w = 11 \), which is correct.

Step4: Test Intervals

The critical points are \( w = -12 \) and \( w = 11 \), dividing the number line into \( (-\infty, -12) \), \( (-12, 11) \), and \( (11, \infty) \). Since width can't be negative, we only consider \( w > 0 \). Test \( w = 0 \) in \( w^2 + w - 132 \): \( 0 + 0 - 132 = -132 < 0 \). Test \( w = 12 \): \( 144 + 12 - 132 = 24 > 0 \). So the inequality \( w^2 + w - 132 \geq 0 \) holds when \( w \leq -12 \) or \( w \geq 11 \). But since width \( w > 0 \), we take \( w \geq 11 \).

Step5: Solution Set

If length \( L = w + 1 \), then when \( w \geq 11 \), \( L \geq 12 \). So the width should be at least \( 11 \) units, and length at least \( 12 \) units.

Answer:

The width \( w \geq 11 \) (and length \( L = w + 1 \geq 12 \)) after correcting the quadratic inequality to \( w^2 + w - 132 \geq 0 \).