Answer:
[tex]126in^2=2w^2-36w[/tex] and w= 21 inches.
Step-by-step explanation:
The placemat has an area of [tex]126in^2[/tex]. Now, let's call w to the width, the exercise says that the length of the placemat is four times the quantity of nine less than half its width, that is
length = 4(w/2-9).
Now, the area is length*width, so
[tex]A=w*4(\frac{w}{2}-9)= \frac{4w^2}{2}-36w = 2w^2-36w.[/tex]
and we know that [tex]A=126in^2[/tex], so
[tex]126in^2=2w^2-36w.[/tex]
Now, let's find the solution.
[tex]2w^2-36w-126=0[/tex] is a quadratic formula of the form [tex]ax^2+bx+c[/tex] with a=2, b= -36 and c = -126. The solutions will be
[tex]w=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
[tex]w=\frac{-(-36)\pm\sqrt{36^2-4(2)(-126)}}{2a}[/tex]
[tex]w=\frac{36\pm\sqrt{1296+1008}}{2*2}[/tex]
[tex]w=\frac{36\pm\sqrt{2304}}{4}[/tex]
[tex]w=\frac{36\pm48}{4}[/tex]
[tex]w=\frac{36+48}{4}[/tex] or [tex]w=\frac{36-48}{4}[/tex]
[tex]w=\frac{84}{4}[/tex] or [tex]w=\frac{-12}{4}[/tex]
as we are searching a distnace, we are going to use the positive solution, that is
[tex]w=\frac{84}{4}= 21.[/tex]
Then, the width is 21 inches.
