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

We start with the equation \(560=(2x + 3)x\). Expanding the right - hand side using the distributive property \(a(b + c)=ab+ac\) (here \(a = x\), \(b = 2x\), \(c = 3\)), we get \(560 = 2x^{2}+3x\).

Step2: Rearrange into standard quadratic form

Subtract 560 from both sides of the equation to get the quadratic equation in standard form \(ax^{2}+bx + c = 0\). So, \(2x^{2}+3x-560 = 0\).

Step3: Solve the quadratic equation

For a quadratic equation \(ax^{2}+bx + c = 0\), the quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Here, \(a = 2\), \(b = 3\), and \(c=- 560\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(3)^{2}-4\times2\times(-560)=9 + 4480=4489\).
Then, find the square root of the discriminant. \(\sqrt{4489} = 67\) (since \(67\times67 = 4489\)).
Now, substitute into the quadratic formula:
\(x=\frac{-3\pm67}{2\times2}\)
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 discard \(x=-17.5\).

Answer:

The width is \(16\) yd.