Question

the length of a rectangle is 4 more than the width. the area of the rectangle is 60 square yards. what is the length of the rectangle? let w = width and 4 + w = length. which equation represents the situation? w(4 + w) = 60 w + 4 + w = 60 2w + 2(4 + w) = 60 what is the length of the rectangle? 10 yards

Explanation:

Step1: Set up the area - equation

The area formula for a rectangle is $A = length\times width$. Given width $w$ and length $w + 4$ and $A=60$, we get $w(w + 4)=60$. Expanding using the distributive property, we have $w^{2}+4w=60$, or $w^{2}+4w - 60=0$.

Step2: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ (here $a = 1$, $b = 4$, $c=-60$), we can use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. First, calculate the discriminant $\Delta=b^{2}-4ac=(4)^{2}-4\times1\times(-60)=16 + 240=256$. Then $w=\frac{-4\pm\sqrt{256}}{2}=\frac{-4\pm16}{2}$. We have two solutions for $w$: $w_1=\frac{-4 + 16}{2}=\frac{12}{2}=6$ and $w_2=\frac{-4-16}{2}=\frac{-20}{2}=-10$. Since the width cannot be negative, $w = 6$.

Step3: Find the length

Since length $l=w + 4$, substituting $w = 6$ into the formula, we get $l=6 + 4=10$.

Answer:

10 yards