Algebra II math problem on Quadratics. Please help! Thanks!

Question

You 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?

If possible, please include steps. Thank you so much!

Answer 1

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

Answer 2

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