The area of a rectangle is given by: 5x^2 -22x + 8. What are the possible dimensions of the rectangle?

1 answer

A = L ∙ W = 5 x² - 22 x + 8

For any quadratic equation:

a x² - b x + c = a ( x - x₁ ) ( x - x₂ )

where x₁ = x₂ are roots of that quadratic equation.

In this case:

5 x² - 22 x + 8 = 0

a = 5 , b = - 22 , c = 8

The solutions are:

2 / 5 and 4

So

5 x² - 22 x + 8 = 5 ( x - 2 / 5 ) ( x - 4 )

Possible dimensions are:

L = 5 ( x - 2 / 5 ) , W = x - 4

L = 5 x - 2 , W = x - 4

and

L = x - 2 / 5 , W = 5 ( x - 4 )

L = x - 2 / 5 , W = 5 x - 20