The length of a rectangular frame is represented by the expression [tex]2x+4[/tex], and the width of the rectangular frame is represented by the expression [tex]2x+10[/tex]. Write an equation to find the width of the rectangular frame that has an initial area of 120 square inches.

A. [tex]2x^2 + 20x - 80 = 0[/tex]

B. [tex]4x^2 + 28x + 40 = 0[/tex]

C. [tex]4x^2 + 28x - 80 = 0[/tex]

D. [tex]x^2 + 8x + 20 = 0[/tex]



Answer :

To write an equation that represents the relationship between the length, width, and area of a rectangular frame with a given initial area of 120 square inches, we start by expressing the length and the width in terms of [tex]\( x \)[/tex].

The length ([tex]\( L \)[/tex]) is given by:
[tex]\[ L = 2x + 4 \][/tex]

The width ([tex]\( W \)[/tex]) is given by:
[tex]\[ W = 2x + 10 \][/tex]

The area ([tex]\( A \)[/tex]) of a rectangle is calculated by multiplying its length and width:
[tex]\[ A = L \times W \][/tex]

Substituting the expressions for the length and width, we get:
[tex]\[ A = (2x + 4) \times (2x + 10) \][/tex]

Next, we need to expand this expression:
[tex]\[ A = (2x + 4)(2x + 10) \][/tex]

Using the distributive property (also known as the FOIL method for binomials):
[tex]\[ A = 2x \cdot 2x + 2x \cdot 10 + 4 \cdot 2x + 4 \cdot 10 \][/tex]
[tex]\[ A = 4x^2 + 20x + 8x + 40 \][/tex]
[tex]\[ A = 4x^2 + 28x + 40 \][/tex]

We are given that the area ([tex]\( A \)[/tex]) is 120 square inches:
[tex]\[ 4x^2 + 28x + 40 = 120 \][/tex]

To convert this into a standard form quadratic equation, we need to set the equation to 0. We do this by subtracting 120 from both sides:
[tex]\[ 4x^2 + 28x + 40 - 120 = 0 \][/tex]
[tex]\[ 4x^2 + 28x - 80 = 0 \][/tex]

Therefore, the correct equation that represents the given conditions is:
[tex]\[ 4x^2 + 28x - 80 = 0 \][/tex]

From the given choices:
1. [tex]\( 2x^2 + 20x - 80 = 0 \)[/tex]
2. [tex]\( 4x^2 + 28x + 40 = 0 \)[/tex]
3. [tex]\( 4x^2 + 28x - 80 = 0 \)[/tex]
4. [tex]\( x^2 + 8x + 20 = 0 \)[/tex]

The correct solution matches the third equation:
[tex]\[ 4x^2 + 28x - 80 = 0 \][/tex]

Thus, the answer is:
[tex]\[ \boxed{3} \][/tex]