Answer to Question #108756 in Algebra for Judy Norelus

Question

The length of a rectangular garden is 3 m greater than the width. The area of the garden is 88 m squared. Find the dimensions of the garden.

Answer 1

Let's assume that the length of the garden is x. The width of the garden will be x - 3. To find the area of a rectangle, we multiply the length of the rectangle by the width of the rectangle. We can write the equation:

x*(x - 3) = 88

x² - 3x = 88

x² - 3x - 88 = 0

a = 1, b = -3, c= -88

$x_1=\dfrac{-b-\sqrt{b^2-4ac}}{2a}$

$x_2=\dfrac{-b+\sqrt{b^2-4ac}}{2a}$

$x_1=\dfrac{-(-3)-\sqrt{(-3)^2-4*1*(-88)}}{2*1}=$

$=\dfrac{3-\sqrt{9+352}}{2}=\dfrac{3-19}{2}=-8$

$x_2=\dfrac{-(-3)+\sqrt{(-3)^2-4*1*(-88)}}{2*1}=$

$=\dfrac{3+\sqrt{9+352}}{2}=\dfrac{3+19}{2}=11$

x₁ = -8 does not satisfy our case because the length of the garden can't be less than zero.

So the only solution, the length of the garden is x = 11(meters).

The width of the garden will be x - 3 = 11 - 3 = 8(meters).

The answer is 11 and 8 meters.