Question

A picture has a width that is 17 inches shorter than the length, and its total area is 480 square inches. Find the dimensions of the picture if x represents the length of the picture
Please Help

Answer 1

What is the formula for the area of a rectangle?

Answer 2

\[A=lw\]

Answer 3

Very good. We know \(l = x\), right? "if x represents the length"

Answer 4

"width that is 17 inches shorter than the length"

If the length is represented by \(x\), how do we write an expression for the width in terms of \(x\)?

Answer 5

\[x ^{2}-17x-480?\]

Answer 6

\[=0\]

Answer 7

so really to write an expression for the width we would have to subtract 34 from 480 and then divide that by 4 correct?

Answer 8

No, if the length is \(x\), and the width is 17 inches shorter than the length, isn't the width \(x-17\)?

Answer 9

Sorry, you skipped ahead and I didn't realize it :-) Yes, \[x^2-17x-480 = 0\]does represent the scenario we have. Now you solve for \(x\), and choose the value which makes sense for this scenario.

Answer 10

\[length=128.5\] \[width= 111.5\]

Answer 11

correct?

Answer 12

Sadly, no. Does 128.5*111.5 = 480?

How did you find those values?

Answer 13

oh wait I forgot to divide lol

Answer 14

\[length=32\] \[width=15\]

Answer 15

correct? and I found the previous values by subtracting 34 from 480 but then I forgot to go through the division

Answer 16

There you go. I'm still not sure how subtracting 34 from 480 is a route to the solution, but you don't have to explain it to me.

\[x^2+17x-480 = 0\]\[a = 1, b = 17, c = -480\]\[x = \frac{-17\pm\sqrt{(17)^2 -4(1)(-480)}}{2(1)} = \frac{1}{2}(-17\pm\sqrt{2209}) = \frac{1}{2}(-17\pm47) \]\[\qquad = -15,\, 32 \]
Only \(x=32\) makes sense as the dimension of a physical object.

Answer 17

Awesome thank you so much @whpalmer4

Answer 18

You're welcome!