Question 1014558

Hi outstanding tutors,
I have a question that is in the quadratic equation solving chapter of my book. It is geometry related, however. Here it is:

Is it possible for a rectangle with a perimeter of 52 centimeters to have an area of 148.75 square centimeters? Explain. A diagram of the rectangle is given with x representing the with and 26 - x representing the length. 
I know that P=2(l+w) and A=wl. I think somehow a quadratic equation is implemented to solve this. Please help! Thanks!!
<pre>From your description length = x
Let width be W
Since Perimeter = 52 cm, we get: 2(x + W) = 52______2(x + W) = 2(26)____x + W = 26______W = 26 - x
testing whether or not 148.75 can be the area, we get: x(26 - x) = 148.75
{{{26x - x^2 = 148.75}}}
{{{x^2 - 26x + 148.75 = 0}}}
Using the quadratic equation formula, completing the square, or factoring (the trinomial would need to contain all INTEGERS), we find that:
x = 17.5 or x = 8.5, and so, the dimensions of this rectangle would be: {{{highlight(highlight_green(highlight(matrix(1,5,17.5, cm, by, 8.5, cm))))}}}, so it's {{{highlight_green(matrix(1,2,DEFINITELY, POSSIBLE))}}}