Answer:
I suspect it can = 0 when the roots are solved for.
x = 10.44 which is the only positive root. There is another one, but I think it gives a minus number for the white area.
Step-by-step explanation:
There are two areas that are possible given the dimensions of the two rectangles. The trick is to find both of them. and compare.
Step One
Find the area of the inside rectangle (unshaded area)
Area = (2x - 1)*x Use the distributive property to find the area
Area = 2x^2 - x
Step Two
Find the area of the outside of the frame
Area = (x + 5)(x + 3) Remove the brackets using FOIL
Area = x^2 + 5x + 3x + 15
Area = x^2 + 8x + 15
Step Three
Find the area of the shaded region
Area shaded = x^2 + 8x + 15 – (2x^2 – x) Remove the brackets
Area shaded = x^2 + 8x + 15 – 2x^2 + x Combine like terms
Area shaded = x^2 + 9x + 15 - 2x^2
Area shaded = -x^2 + 9x + 15
Step Four
Find the roots of the quadratic.
a = -1
b = 9
c = 15
Using the quadratic formula, and substituting for a b and c, we get
[tex]\text{x = }\dfrac{ -b \pm \sqrt{b^{2} - 4ac } }{2a}[/tex]
[tex]\text{x = }\dfrac{ -9 \pm \sqrt{9^{2} - 4*(-1)(15 } }{2(-1)}[/tex]
[tex]\text{x = }\dfrac{ -9 \pm \sqrt{81 + 4*(1)(15 } }{2(-1)}[/tex]
There is only 1 value that works for this problem and that is when x = 10.44
