Question

part 1 of 3
(a) write an equation in terms of x that represents the given relationship.
the equation is 377 = (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 \boxed{} yd.

Explanation:

Step1: Expand the equation

We start with the equation \( 377=(2x + 3)x \). First, we expand the right - hand side using the distributive property \( a(b + c)=ab+ac \). Here, \( a = x \), \( b=2x \) and \( c = 3 \), so \( (2x + 3)x=2x^{2}+3x \). The equation becomes \( 2x^{2}+3x - 377=0 \).

Step2: Use the quadratic formula

For a quadratic equation of the form \( ax^{2}+bx + c = 0 \), the quadratic formula is \( x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \). In our equation \( 2x^{2}+3x - 377 = 0 \), we have \( a = 2 \), \( b = 3 \) and \( c=- 377 \).

First, calculate the discriminant \( D=b^{2}-4ac=(3)^{2}-4\times2\times(-377)=9 + 3016=3025 \).

Then, find the square root of the discriminant \( \sqrt{D}=\sqrt{3025} = 55 \).

Now, substitute into the quadratic formula:
\( x=\frac{-3\pm55}{2\times2}=\frac{-3\pm55}{4} \)

We have two solutions:

• \( x_{1}=\frac{-3 + 55}{4}=\frac{52}{4}=13 \)
• \( x_{2}=\frac{-3-55}{4}=\frac{-58}{4}=-14.5 \)

Since the dimension (width) cannot be negative, we take \( x = 13 \).

Answer:

13