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.

Guest Apr 30, 2017

#1**+3 **

(I think you meant to put cm^{2} )

area = 54

length = 2 + x

width = 2x

area = (length)(width) Plug in values given in the problem.

54 = (2 + x) (2x) Distribute

54 = 4x + 2x^{2} Subtract 54 from both sides and rearrange.

0 = 2x^{2} + 4x - 54 Divide through by 2

0 = x^{2} + 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

hectictar
May 1, 2017