Question

The length of a rectangle is 3yd more than twice the width x. The area is 434 yd^2.
Write an equation in terms of x that represents the given relationship.

Answer 1

To represent the given relationship of the rectangle's dimensions, we first establish the length in terms of the width \( x \).

The length \( L \) is given as:
\[
L = 2x + 3
\]

The area \( A \) of a rectangle is calculated by the formula:
\[
A = L \times W
\]
where \( W \) is the width. In this case, the area is given as \( 434 \, \text{yd}^2 \), so we can write:
\[
A = (2x + 3) \times x
\]

Now, substituting the area into the equation, we have:
\[
434 = (2x + 3) \times x
\]

Expanding the right side:
\[
434 = 2x^2 + 3x
\]

To form a standard quadratic equation, we can rearrange it:
\[
2x^2 + 3x - 434 = 0
\]

This equation \( 2x^2 + 3x - 434 = 0 \) represents the given relationship in terms of \( x \).