xCorrosive

I have a hint that they provided. 

Let $y = x + \frac{a}{3}$ and rewrite $x^3 + ax^2 + bx + c = 0$ in terms of $y$. 

Thanks, CPhill !!

Wait I have a thought. 

There are only two cases in which these two equations are equal to each other.

The first case is when two the variables are $0$, effectively leading the equation to be zero. 

The second case is when $x^2y^2z^2 = 1$. 

We're looking for the square root of this, so does it mean that the only solutions are $-1$ and $1$. Or are there other possible solutions? 

$$\text{cool!}$$

Anyway, I didn't exactly multiply by $xyz$, but now I'm at a point where I have this equation

$(y-x)(y-z)(z-x)= \frac{(y-x)(y-z)(z-x)}{x^2y^2z^2}$

Any pointers on how to continue from here? 

$$wow this works thanks.$$

Thank you! Will try the hint. 

Floor basically means... rounding down to the nearest number. 

Thus, we should set $3 \leq x^2 < 4$. 

Similarly, we can also set $-3 \leq x^2 < -4$. 

Taking the square root gives... $\sqrt{3} \leq x < \sqrt{4}$ and $-\sqrt{3} \leq x < -\sqrt{4}$.

Thus entering in integer notation $(-\sqrt{4}, -\sqrt{3}] \cup [\sqrt{3}, \sqrt{4})$. (square bracket = include end, round bracket = don't include end, U = union (include both in solution). 

Edit: Solved it by myself, no need to help me anymore :)

After doing $\frac{3}{5}$ of his homework he has $\frac{2}{5}$ left. 

Multiplying the $\frac{2}{5}$ or his homework must be 6 questions. This is because $1 - \frac{1}{3} = \frac{2}{3} = 4$, as your problem states "the last 4 questions". Thus, 1/3 is equal to 2 questions, and  $\frac{2}{5}$ or his homework must be 6 questions. Multiply $6 \cdot \frac{5}{2} = 15$. 

Therefore, he was assigned 15 questions in total for the weekend. 

The answer is $390$, I can't think of a good way to do this than just expanding all of it. 

Expanding all of these two expressions gives $2x^4+4097y^4+36x^3y+390x^2y^2 + 2052xy^3$. From this you can clearly see that the coefficient of $x^2y^2 = 390$.

Sorry about that!

I meant to say AP^2 + BQ^2 = PQ^2