Question

the area of a rectangle is 54 m², and the length of the rectangle is 3 m more than twice the width. find the dimensions of the rectangle.

Explanation:

Step1: Let the width be $x$ meters.

The length is $2x + 3$ meters. The area formula for a rectangle is $A=\text{length}\times\text{width}$. So we have the equation $x(2x + 3)=54$.

Step2: Expand the left - hand side.

Expand $x(2x + 3)$ to get $2x^{2}+3x=54$. Rearrange it to the standard quadratic form $2x^{2}+3x - 54 = 0$.

Step3: Solve the quadratic equation.

For a quadratic equation $ax^{2}+bx + c = 0$ ($a = 2$, $b = 3$, $c=-54$), we can use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. First, calculate the discriminant $\Delta=b^{2}-4ac=(3)^{2}-4\times2\times(-54)=9 + 432 = 441$. Then $x=\frac{-3\pm\sqrt{441}}{2\times2}=\frac{-3\pm21}{4}$.
We get two solutions for $x$: $x_1=\frac{-3 + 21}{4}=\frac{18}{4}=4.5$ and $x_2=\frac{-3 - 21}{4}=\frac{-24}{4}=-6$. Since the width cannot be negative, we take $x = 4.5$ meters.

Step4: Find the length.

Substitute $x = 4.5$ into the length formula $l=2x + 3$. So $l=2\times4.5+3=9 + 3=12$ meters.

Answer:

Length: 12 m
Width: 4.5 m