Inspirational

No one got an answer? 

Is Alan's answer good enough?

or do you just want the answer? 

Yep you are correct

d is definitely not it

\(\frac{d}{dx}\left(e^{3x}\sin \left(3x\right)\right)\\ \frac{d}{dx}\left(e^{3x}\right)\sin \left(3x\right)+\frac{d}{dx}\left(\sin \left(3x\right)\right)e^{3x}\\ \frac{d}{dx}\left(e^{3x}\right)=e^{3x}\cdot \:3\\ \frac{d}{dx}\left(\sin \left(3x\right)\right)=\cos \left(3x\right)\cdot \:3\\ e^{3x}\cdot \:3\sin \left(3x\right)+\cos \left(3x\right)\cdot \:3e^{3x}\\\)

Hmm...

That's confusing

Report an error if possible

You are right, I think.

I sense more AOPS

Realize that 7^13=96889010407.

therefore your answer is 13

Well A and B are our candidates

but the thing is that a will be a positive decimal whereas b will be >1

so b

\(x = {-4 \pm \sqrt{4^2-4(3)(12)} \over 2(3)}\\ x = {-4 \pm \sqrt{-128} \over 6}\\ x = {-4 \pm 3\sqrt{-2} \over 6}\\ ({-4 + 3\sqrt{-2} \over 6})^2+({-4 - 3\sqrt{-2} \over 6})^2\\ {(-4 + 3\sqrt{-2})^2 \over 6^2}+{(-4 - 3\sqrt{-2})^2 \over 6^2}\\ {(-4 + 3\sqrt{-2})^2 \over 36}+{(-4 - 3\sqrt{-2})^2 \over 36}\\ {(-4 + 3\sqrt{-2})^2+(-4 - 3\sqrt{-2})^2 \over 36}\\\)

Well this is kind of bashy, lOl.

\(r + s = -\frac{4}{3} \text{ due to vieta}\\ (r + s)^2 = \frac{16}{9} \text{ because we square -4}\\ r^2+s^2+2rs=\frac{16}9\\ rs=4 \text{ due to vieta}\\ r^2 + s^2 = \frac{16}{9} - 2(4) = \frac{16}{9}-8=-6\frac{2}{9}\)

Someone check this

Ahh, I see how you're doing it. Yes, guest, you can also do it in cents or just multiply by a 100 if you want cleaner numbers to work with. Nice catch!