Question

the area of the conference table in mr. nathans office must be no more than 175 ft². if the length of the table is 18 ft more than the width, x, which interval can be the possible widths?
○ 0 < x ≤ 25
○ 0 < x ≤ 7
○ 0 ≤ x ≤ 7
○ 0 ≤ x ≤ 25

Explanation:

Step1: Define variables and area formula

Let width be \( x \), length is \( x + 18 \). Area of rectangle: \( A = x(x + 18) \).

Step2: Set up inequality

Area \( \leq 175 \), so \( x(x + 18) \leq 175 \). Expand: \( x^2 + 18x - 175 \leq 0 \).

Step3: Solve quadratic equation

Solve \( x^2 + 18x - 175 = 0 \). Using quadratic formula \( x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \), \( a = 1 \), \( b = 18 \), \( c=-175 \).
Discriminant: \( 18^2 - 4(1)(-175)=324 + 700 = 1024 \), \( \sqrt{1024}=32 \).
Roots: \( x=\frac{-18\pm32}{2} \). \( x_1=\frac{-18 + 32}{2}=7 \), \( x_2=\frac{-18 - 32}{2}=-25 \).

Step4: Analyze inequality and domain

Quadratic \( x^2 + 18x - 175 \) opens up (since \( a>0 \)). Inequality \( x^2 + 18x - 175 \leq 0 \) holds between roots: \( -25 \leq x \leq 7 \). But width \( x>0 \) (can't be zero or negative), so \( 0 < x \leq 7 \).

Answer:

\( 0 < x \leq 7 \) (corresponding to the option: \( 0 < x \leq 7 \))