Question

the area of a rectangle is 16 in². the length is 3 times the width plus 2. use w for width. what is the width in inches?

Explanation:

Step1: Define length in terms of width

Let width be \( w \). Length \( l = 3w + 2 \).

Step2: Use area formula for rectangle

Area of rectangle \( A = l \times w \). Given \( A = 16 \), so \( (3w + 2)w = 16 \).

Step3: Expand and form quadratic equation

Expand: \( 3w^2 + 2w = 16 \). Rearrange: \( 3w^2 + 2w - 16 = 0 \).

Step4: Solve quadratic equation

Use quadratic formula \( w=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = 2 \), \( c = -16 \).
Discriminant \( D = 2^2 - 4\times3\times(-16)=4 + 192 = 196 \).
\( \sqrt{D}=\sqrt{196}=14 \).
\( w=\frac{-2\pm14}{6} \).
Two solutions: \( w=\frac{-2 + 14}{6}=\frac{12}{6}=2 \) and \( w=\frac{-2 - 14}{6}=\frac{-16}{6}=-\frac{8}{3} \).
Since width can't be negative, we take \( w = 2 \).

Answer:

The width is \( 2 \) inches.