SOLUTION: The frame around a picture measures x inches across. The dimension of the picture frame is 28 inches by 32 inches. If the area of the picture is 192 square inches, find the dimensi

Question 1170832: The frame around a picture measures x inches across. The dimension of the picture frame is 28 inches by 32 inches. If the area of the picture is 192 square inches, find the dimensions of the picture.
Found 3 solutions by CubeyThePenguin, ikleyn, greenestamps:
Answer by CubeyThePenguin(3113)   (Show Source): You can put this solution on YOUR website!
(28 - 2x)(32 - 2x) = 192

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=5008 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 18.230602004318, 6.43606466234862. Here's your graph:

Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.

(28-2x)*(32-2x) = 192 (the area of the picture, in square inches) (14-x)*(16-x) = 48 x^2 - 30x + 14*16 - 48 = 0 x^2 - 30x + 176 = 0. Factor it <<<---=== re-edited starting from this point (x-8)*(x-22) = 0. The roots are x= 8 and x= 22. Due to the problem's meaning, only the root x= 8 is the possible width of the frame (22 is not). Thus the dimensions of the picture are 28-2*8 = 12 inches and 32 - 2*8 = 16 inches. ANSWER

Solved.

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To see many other similar solved problems, look into the lessons
    - Problems on the area and the dimensions of a rectangle surrounded by a strip
    - Cynthia Besch wants to buy a rug for a room
    - Problems on a circular pool and a walkway around it
    - OVERVIEW of lessons on dimensions and the area of rectangles and circles and their elements
in this site.

Also, you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lesson is the part of this online textbook under the topic
"Dimensions and the area of rectangles and circles and their elements".

Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

This problem can be solved in a few seconds informally, with just a bit of mental arithmetic.

One response you have received presumably used an automatic problem solver; but something is wrong somewhere in the calculations, and the solution is irrational and therefore only approximate.

The other response fails to see the solution, after setting up the problem correctly.

But the dimensions of the picture are "nice' numbers; here is the quick and easy informal solution....

The difference between the length and width of the frame (including picture) is 4 inches; since the frame around the picture is of uniform width, the difference between the length and width of the picture is 4 inches.

So to solve the problem, we only need to find two numbers with a difference of 4 and a product of 192. Of course you could solve THAT problem algebraically; but some quick mental arithmetic finds the numbers to be 12 and 16.

So the picture is 12 inches by 16 inches; since the picture with frame is 28 inches by 32 inches, the width of the frame is 8 inches.

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