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