Ken wants to build a table and put a border around it. The table and border must have an area of 3,276 square feet. The table is 36 inches wide and 72 inches long without the boarder. Which quadratic equation can be used to determine the thickness of the border, x?

4x2 + 216x + 2,592 = 0

4x2 + 216x - 684 = 0

2x2 + 216x - 3,276 = 0

x2 + 108x + 3,276 = 0

A = 3.276 ft²
A = ( W + 2 x ) · ( L + 2 x )
( 36 + 2 x ) · ( 72 + 2 x ) = 3,276
2,592 + 72 x + 144 x + 4 x² = 3,276
4 x² + 216 x + 2,592 - 3,276 = 0
Answer:
B ) 4 x² + 216 x - 684 = 0

Answer Link

calculista calculista · Answer 2 · 2018-12-31T18:18:06+01:00

Answer:

[tex]4x^{2}+216x-684=0[/tex]

Step-by-step explanation:

case A) The dimensions of the table are in inches

Let

x------> the thickness of the border

we know that

The table and border must have an area of [tex]3,276[/tex] square feet

Convert the dimensions of the table to feet

remember that

[tex]1\ ft=12\ in[/tex]

[tex]36\ in=36/12=3\ ft[/tex]

[tex]72\ in=72/12=6\ ft[/tex]

so

[tex]3,276=(6+2x)*(3+2x)[/tex]

Solve for x

[tex]3,276=6*3+6*2x+2x*3+4x^{2} \\3,276=18+12x+6x+4x^{2} \\4x^{2}+18x+18-3,276=0\\4x^{2}+18x-3,258=0[/tex]

case B) The dimensions of the table are in feet

Let

x------> the thickness of the border

we know that

The table and border must have an area of [tex]3,276[/tex] square feet

so

[tex]3,276=(72+2x)*(36+2x)[/tex]

Solve for x

[tex]3,276=72*36+72*2x+2x*36+4x^{2} \\3,276=2,592+144x+72x+4x^{2} \\4x^{2}+216x+2,592-3,276=0\\4x^{2}+216x-684=0[/tex]