austinruffino

It's alright, and sorry about mixing up the signs!

\(f(x) = -x^2-6x+5\)

\(f(x) = {-(-6) \pm \sqrt{(-6)^2-4(-1))5)} \over 2(-1)}\)

\(f(x)=-3 \pm \sqrt{14}\)

\(\bf f(x) \approx -6.742, .742\)

Because the two cylinders are similar, you can use a proportion to determine the height of the larger cylinder

\(3/7 = 5/x\)

By solving the proportion, you get a value for x

\(x = 1 \frac23\)

Now, by multiplying the radius of the larger cylinder by the x value we just found, we can find the height of the larger cylinder

\(1 \frac23*7= 11 \frac23\)

Now, just subsitute the newly found height, and width of the cylinder into the surface area equation

\(A = 2 \pi r h + 2 \pi r^2\)

\(A = 2 \pi (5)(11\frac23)+2 \pi(5)^2\)

\(A \approx 523.6\)

The surface area of the larger cylinder is \(\textbf 5 \textbf 2 \textbf 3 \textbf . \textbf 6 \space \textbf m^ \textbf 2\)

Sorry Austin, I thought I was deleting my own incorrect answer.

I can't put it back easily. Sorry.

Your answer was mostly good but it did have an error in it.   :(

\(y= -6x^2+42x -6\)

First, factor out a -6

\(y = -6(x^2 - 7x + 1)\)

Then, set the value in the parentheses equal to 0

\(x^2-7x+1=0\)

Now, use the quadratic formula to solve

\(y = {(7) \pm \sqrt{(7)^2-4(1)(1)} \over 2(1)}\)

\(y = {7 \pm 3 \sqrt5 \over 2}\)

\(y \approx 6.854, 0.146\)

Your zeros are approximately 6.854, and 0.146