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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
+7324
+3

(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

hectictar  May 1, 2017
#2
+7024
+1

(x + 2)(2x) = 54

2x^2 + 4x = 54

2x^2 + 4x - 54 = 0

x^2 + 2x - 27 = 0

(x + 1)^2 = 28

x + 1 = +- sqrt(28)

x = -1 +- 2sqrt(7)

x = 2sqrt7 - 1 OR -1 - 2sqrt7(rejected)

x = 2sqrt7 - 1

MaxWong  May 1, 2017