Question

The length of a rectangle is 5 yd more than twice the width x. The area is 462 yd".

Answer 1

The given problem is about rectangle areas.

The area of a rectangle is defined as

$A=w\cdot l$

Where w is width and l is the length.

In this case, we know that the length is 5 yards more than twice the width, where the last one is expressed as x.

$l=2x+5,w=x$

We also know by given that

$A=462yd^2$

Using all the given information, we have

$462=x(2x+5)$

Where we solve for x, first, we apply the distributive property.

$462=2x^2+5x$

Now, we move all terms to one side only

$2x^2+5x-462=0$

Using a calculator, we have

$x_1=14,x_2=-\frac{33}{2}$

Where 14 represents the width because distances cannot be expressed by negative numbers.

$l=2(14)+5=28+5=33$

Therefore, the width of the rectangle is 14 yards, and its length is 33 yards.