Question

1. the length of one side of a rectangular pool is three feet more than twice the width. the area of the pool is 527 square feet. use a quadratic equation to find the dimensions of the pool. show all work. (4 pts)

\td. write system of equations: ________________________ (equation 1)
\t\t\t\t\t\t\t\t\t\t________________________ (equation 2)
\te. write the quadratic equation in term of one variable:
\t___________________________________________ (equation 3)
\tf. find 2 sides (length and width):

answer(s): _____________________

Explanation:

Part d: Write system of equations

Step 1: Define variables

Let \( w \) be the width of the rectangular pool (in feet) and \( l \) be the length of the rectangular pool (in feet).

Step 2: Write Equation 1 (length in terms of width)

The length is three feet more than twice the width. So, \( l = 2w + 3 \).

Step 3: Write Equation 2 (area of rectangle)

The area \( A \) of a rectangle is given by \( A = l \times w \). We know the area is 527 square feet, so \( lw = 527 \).

Part e: Write the quadratic equation in term of one variable

Step 1: Substitute \( l \) from Equation 1 into Equation 2

Substitute \( l = 2w + 3 \) into \( lw = 527 \). We get \( (2w + 3)w = 527 \).

Step 2: Expand and simplify

Expand the left - hand side: \( 2w^{2}+3w = 527 \).
Subtract 527 from both sides to get the quadratic equation in standard form: \( 2w^{2}+3w - 527=0 \).

Part f: Find 2 sides (length and width)

Step 1: Solve the quadratic equation \( 2w^{2}+3w - 527 = 0 \)

We can use the quadratic formula \( w=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \) for a quadratic equation \( ax^{2}+bx + c = 0 \). Here, \( a = 2 \), \( b = 3 \), and \( c=-527 \).

First, calculate the discriminant \( \Delta=b^{2}-4ac \):
\( \Delta=(3)^{2}-4\times2\times(-527)=9 + 4216=4225 \)

Then, find \( w \):
\( w=\frac{-3\pm\sqrt{4225}}{2\times2}=\frac{-3\pm65}{4} \)

We have two solutions for \( w \):

• \( w=\frac{-3 + 65}{4}=\frac{62}{4}=15.5 \)
• \( w=\frac{-3-65}{4}=\frac{-68}{4}=-17 \)

Since the width cannot be negative, we take \( w = 15.5 \) feet.

Step 2: Find the length

Use \( l = 2w+3 \). Substitute \( w = 15.5 \) into this equation:
\( l=2\times15.5 + 3=31 + 3=34 \) feet.

Answer:

s:

Part d:

Equation 1: \( \boldsymbol{l = 2w + 3} \)

Equation 2: \( \boldsymbol{lw=527} \)

Part e:

Equation 3: \( \boldsymbol{2w^{2}+3w - 527 = 0} \)

Part f:

Width \( w=\boldsymbol{15.5} \) feet, Length \( l=\boldsymbol{34} \) feet.