Question 378469
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Let *[tex \Large x] represent the width of the photo.  Then the length of the photo must be *[tex \Large x\ +\ 10].  Since the frame is 2 cm wide, it adds 2 cm to EACH side, top, and bottom.  So the overall width, including the frame, must be *[tex \Large x\ +\ 4] cm, and the overall length must be *[tex \Large x\ +\ 14] cm.  The area of a rectangle is the length times the width:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 4)(x\ +\ 14)\ =\ 112]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 18x\ +\ 56\ -\ 112\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 18x\ -\ 56\ =\ 0]


Solve the quadratic, exclude the negative root, and the positive root is the width of the photo.  Note:  This ugly beast does NOT factor.  You need to haul out the quadratic formula.  And this is one <i>really</i> odd shaped photo.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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