The length of a rectangle is 10m more than twice the width. The

area is 120 m². What are the dimensions of the rectangle to the

nearest tenth?

5.6, 21.3

Step-by-step explanation:

Let x be the width.

We know that the length is 10+2x. The area of a rectangle is length times width, so substitute the values.

x * (10+2x) = 120

10x+2x^2=120

Divide both sides by 2.

x^2+5x=60

Add 25/4 to both sides so we can make a square.

x^2+5x + 25/4 = 60+25/4=265 / 4

Factor the left side.

(x + 5 / 2) ^2 = 265 / 4

x+2.5 = (positive or negative) sqrt (265/4)

x = (positive or negative) sqrt (265/4) - 2.5

But, we know that if the equation is negative, the width of the rectangle would be negative. Therefore, x=positive sqrt (265/4) - 2.5

Then, we will find the sides.

sqrt66.25-2.5=5.63941, or about 5.6

10 + (2 (sqrt66.25-2.5)) = 21.278821, or about 21.3