The area of a rectangle is 54 cm. The length is 2 cm more than an x and the width is twice x. Solve for x. Round to the nearest whole number.
+9466 (I think you meant to put cm2 )
area = 54
length = 2 + x
width = 2x
area = (length)(width) Plug in values given in the problem.
54 = (2 + x) (2x) Distribute
54 = 4x + 2x2 Subtract 54 from both sides and rearrange.
0 = 2x2 + 4x - 54 Divide through by 2
0 = x2 + 2x - 27 Use the quadraic formula to solve for x.
\(x = {-2 \pm \sqrt{2^2-4(1)(-27)} \over 2(1)}= {-2 \pm 4\sqrt{7} \over 2}=-1\pm 2\sqrt7\)
x ≈ 4.292 or x ≈ -6.292
If the length is 2 cm more than x, the question probably wants the positive value for x. To the nearest whole numer that is x = 4 cm
