Question

part 1 of 3
(a) write an equation in terms of x that represents the given relationship.
the equation is 560 = (2x + 3)x.
part: 1 / 3
part 2 of 3
(b) solve the equation to find the dimensions of the given shape.
the width is \\(\square\\) yd.

Explanation:

Step1: Expand the equation

Given the equation \( 560=(2x + 3)x \), expand the right - hand side:
\( 560 = 2x^{2}+3x \)
Rearrange it to the standard quadratic form \( ax^{2}+bx + c = 0 \):
\( 2x^{2}+3x - 560=0 \)

Step2: Solve the quadratic equation

For a quadratic equation \( ax^{2}+bx + c = 0 \) (here \( a = 2 \), \( b = 3 \), \( c=- 560 \)), we can use the quadratic formula \( x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \) or try to factor the quadratic.
First, calculate the discriminant \( D=b^{2}-4ac=(3)^{2}-4\times2\times(-560)=9 + 4480 = 4489 \)
Then, \( x=\frac{-3\pm\sqrt{4489}}{4} \)
We know that \( \sqrt{4489}=67 \) (since \( 67\times67 = 4489 \))
So \( x=\frac{-3\pm67}{4} \)
We have two solutions:
\( x_{1}=\frac{-3 + 67}{4}=\frac{64}{4}=16 \)
\( x_{2}=\frac{-3-67}{4}=\frac{-70}{4}=-17.5 \)
Since the dimension (width) cannot be negative, we take \( x = 16 \)

Answer:

The width is \( \boldsymbol{16} \) yd.