Stephen R. answered • 10/02/17

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From the information given you can derive the equation.

l = 2w-3 and A= l x w = 27 ft

^{2}27 ft2 = (2w-3) x w substitute with value of l in terms of w

2w

^{2}-3w -27 = 0 standard quadratic equation form ax^{2}+ bx + c = 0the roots are:

w = [- b +/- sqr(b

^{2}-4ac)] / 2aw = [- (-3) +/- sqr((-3)

^{2}-4(2)(-27))] / 2(2)w = [3 +/- sqr(9 +216)] / 4

w = [3 +/- sqr(225)] / 4

w = [3 +/- 15] / 4

w = 18 / 4 and w = 12 / 4 = 3

solve for l

l = 2w - 3 = 2(18/4) - 3 = 6 or l = 2w - 3 = 2(3) - 3 = 3

the two roots are 18/4 and 3

verify

l x w = 27

6 x 18/4 = 27

27 = 27 so the root w=18/4 is valid

l x w = 27

3 x 3 ≠ 27

9 ≠ 27 so the root w=3 is not valid

3 x 3 ≠ 27

9 ≠ 27 so the root w=3 is not valid