Question 610249
x^2 = (2x-3)(x-3)


x^2 = 2x^2 - 6x - 3x + 9


0 = 2x^2 - 6x - 3x + 9 - x^2


0 = x^2 - 9x + 9


x^2 - 9x + 9 = 0


Now use the quadratic formula to solve for x


x = (-b+-sqrt(b^2-4ac))/(2a)


x = (-(-9)+-sqrt((-9)^2-4(1)(9)))/(2(1))


x = (9+-sqrt(81-(36)))/(2)


x = (9+-sqrt(45))/2


x = (9+sqrt(45))/2 or x = (9-sqrt(45))/2


x = (9+3*sqrt(5))/2 or x = (9-3*sqrt(5))/2


x = 7.85410196624969 or x = 1.14589803375031


Since each dimension is positive, this means that the only solution is approximately x = 7.85410196624969


Perimeter of the rectangle

P = 2L + 2W


P = 2(2x-3) + 2(x-3)


P = 2(2*7.85410196624969-3) + 2(7.85410196624969-3)


P = 35.1246117974981


So the perimeter of the rectangle is approximately 35.1246117974981 units

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