Online Course Discussion Forum
MC II-A 9.13 and more (Summer:Algebra)
On 9.13, I think the answer should be expressed as A +- sqrt(B), if not, I'm probably getting it wrong. If I'm wrong, could I use a hint?
On 9.21, I've worked into the problem a bit, but I still don't know where to go. Could you perhaps get me started?
9.23: Do I have to do a certain "something" to make it "using Vieta's"? I got the answer from brute force, but I don't know how to get the sum of the dimensions just from the cubic.
I also need help on 9.6, then I should be able to get 9.26.
9.27(b): I got (a), but does b have something to do with the a^3 + b^3 + c^3 - 3abc factorization?
Also I need a boost on 9.28 and 9.29.
Sorry if I'm asking for too much..
For 9.13, you are correct. This should now be fixed.
For the more:
9.21: If $r > s$ are the roots, we know $r+s = -p$, $rs=1$, and $r-s = 1$. Can you use some of these to solve for $r$ and $s$?
9.23: The "using Vieta's" part comes from setting up the cubic you need to solve, not solving the cubic (which you need to do by some brute force).
For 9.6 and 9.26, try making the substitution $u = x+y$ and $v = xy$.
For 9.27: In part (a) you should have $P(z) = 1\cdot z^3 + 0\cdot z^2 + (-3xy)\cdot z + (x^3+y^3)$. Hence in part (b) you want to show $P( -(x+y) ) = 0 $ (that is, substituting $z = -(x+y)$ simplifies to $0$) so $-(x+y)$ is a "zero" of the polynomial $P(z)$.
9.28 is tricky. As a hint, we know $t = -(a+b)(b+c)(a+c)$ (be sure you can explain why). Compare this to the expression $(ab+bc+ac)(a+b+c)$. (Remember we are trying to write everything as expressions we know using Vieta's.)
For 9.29: It is easy to calculate $x +y = 6$ and $xy = 4$. Use this to your advantage.
Hope these help!
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