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Tabitha wants to hang a painting in a gallery. The painting and frame must have an area of 58 square feet. The painting is 7 feet wide by 8 feet long. Which quadratic equation can be used to determine the thickness of the frame, x?
Image of a frame 7 feet wide by 8 feet long, with an of x on the bottom and right sides
x2 + 15x − 2 = 0
x2 + 15x + 58 = 0
4x2 + 30x − 2 = 0
4x2 + 30x + 58 = 0

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you here?

- anonymous

Hmmm consider the drawing
|dw:1442693956662:dw|
We are given the combined area(in respective units)
\[A_{outer}=58\]
and we can find the inner area
\[A_{inner}=7 \times 8=56\]
then the area of the thickness is
\[A_{thickness}=58-56=2\]
Now to find an expression for the area of the thickness in terms of x, we can do
|dw:1442694478789:dw|
This way we have divided out middle section into 4 trapeziums, out of which the left right trapezium are the same and up the down are the same, so they can be just doubled,
Now for first pair of trapezium |dw:1442694716457:dw|
So we have
\[A_{trap1}=\frac{a+b}{2}.h=\frac{8+8+2x}{2}.x=\frac{16+2x}{2}.x=(8+x).x=x^2+8x\]
So we have
\[2 \times A_{trap1}=2x^2+16x\]
Similarly for other pair of trapezium we can do
\[2 \times A_{trap2}=(a+b).h=(7+7+2x).x=(14+2x)x=2x^2+14x\]
So total Area of thickness or the middle section
\[A_{thickness}=2x^2+16x+2x^2+14x=4x^2+30x\]
But we know that
\[A_{thickness}=2\]\[\therefore 4x^2+30x=2 \implies 4x^2+30x-2=0\]

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- anonymous

|dw:1442695223716:dw|

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ok

- anonymous

or you can simply do
\[(8+2x)(7+2x)=58\]\[\implies 4x^2+30x+56=58 \therefore 4x^2+30x-2=0\]

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ty! can oyu help w/ a few mor

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