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Sagot :
This question is solved applying the formula of the area of the rectangle, and finding it's width. To do this, we solve a quadratic equation, and we get that the cardboard has a width of 1.5 feet.
Area of a rectangle:
The area of rectangle of length l and width w is given by:
[tex]A = wl[/tex]
w(2w + 3) = 9
From this, we get that:
[tex]l = 2w + 3, A = 9[/tex]
Solving a quadratic equation:
Given a second order polynomial expressed by the following equation:
[tex]ax^{2} + bx + c, a\neq0[/tex].
This polynomial has roots [tex]x_{1}, x_{2}[/tex] such that [tex]ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})[/tex], given by the following formulas:
[tex]x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}[/tex]
[tex]x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}[/tex]
[tex]\Delta = b^{2} - 4ac[/tex]
In this question:
[tex]w(2w+3) = 9[/tex]
[tex]2w^2 + 3w - 9 = 0[/tex]
Thus a quadratic equation with [tex]a = 2, b = 3, c = -9[/tex]
Then
[tex]\Delta = 3^2 - 4(2)(-9) = 81[/tex]
[tex]w_{1} = \frac{-3 + \sqrt{81}}{2*2} = 1.5[/tex]
[tex]w_{2} = \frac{-3 - \sqrt{81}}{2*2} = -3[/tex]
Width is a positive measure, thus, the width of the cardboard is of 1.5 feet.
Another similar problem can be found at https://brainly.com/question/16995958
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