How to solve this please?
- Thread starter Am Fr
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Harry_the_cat
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Is that the total question? Seems like some info is missing.
HallsofIvy
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I suspect that they want you to factor the quadratic and then assume that the factors give the lengths of two sides. But there can exist infinitely many rectangles with the same area.
For example, if I were told a rectangle had area "6", I might assume that, since 6= 2 x 3 that the rectangle is "2 by 3" so had perimeter 4+ 6= 10. But 6 is also equal to 6= 1 x 6 so another possibility is rectangle with perimeter 2+ 12= 14. And, of course, \(\displaystyle \sqrt{6}\times \sqrt{6}\) so the rectangle might also be a square with perimeter \(\displaystyle 4\sqrt{6}.\)
For example, if I were told a rectangle had area "6", I might assume that, since 6= 2 x 3 that the rectangle is "2 by 3" so had perimeter 4+ 6= 10. But 6 is also equal to 6= 1 x 6 so another possibility is rectangle with perimeter 2+ 12= 14. And, of course, \(\displaystyle \sqrt{6}\times \sqrt{6}\) so the rectangle might also be a square with perimeter \(\displaystyle 4\sqrt{6}.\)