WyzAnt.comYou place a mat around a 25 inch by 21 inch painting. The width of the painting is (25 in + 4x) and the length is (21in + 2x). The area of the mat is 714 sq inches. What is x?
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Algebra II math problem on Quadratics. Please help! Thanks!
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2 Answers
Now revised since comment below so disregard that comment.
I am assuming that you mean the width and length of the painting with the mat already placed around it is 25+4x and 21+2x inches respectively.
You probably know that the area of a rectangle is length times width.
In this case it's just (25+4x)(21+2x).
We are given that this is (714+21*25) in2 so we can equate the two quantities:
(25+4x)(21+2x)=21*25+714
(25+4x)(21+2x)-21*25-714=0
Expand it:
8x2+134x-714=0
Use the quadratic formula or factor:
It factors as 2(x+21)(4x-17)=0
since x>0, x=17/4=4.25 inches
You might be more comfortable expressing the sides after the mat has been added as (4x +25) and (2x + 21). To solve the problem, you know the area of the mat. However, the painting size has to be subtracted from the area of the mat because a mat is hollow in the middle.
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The area of the mat including the middle is (4x +25) (2x + 21) or 8 x2 + 84x + 50x + 525
The area of the painting is (25)(21) which equals 525. You need to subtract that from the expression above to get the 714.
8x + 134x - 714 = 0 2(4x^2 + 67x - 357) = 0 2(x+21)(4x-17) = 0 4x - 17 = 0 4x = 17 x = 17/4 or 4.25
There is another answer which can be obtained for this equation using the factor x + 21 = 0 so x = -21 - Mary K. yesterday
Comments
That is, x is 1.3082 inches
- Roman C. 10/21/2012Nevermind the equation should be (25+4x)(21+2x)=21*25+714=1239.
Revising the above:
The quadratic will be 8x2+134x-714=0 which factors as 2(x+21)(4x-17)=0 so we get x=4/17 inches
- Roman C. 10/21/2012