mathwannabe
Junior Member
- Joined
- Feb 20, 2012
- Messages
- 122
Hello everybody 
I'm really stuck with one problem. Here it goes:
1) \(\displaystyle x1\) and \(\displaystyle x2\) are solutions of the equation \(\displaystyle x^2-x+m^2+2m-3=0\). Value of the real parameter \(\displaystyle m\), for which the sum \(\displaystyle x1^3+x2^3\) is largest, belongs to which set (here they give a bunch of answers, one of which is the correct one).
Since I don't really understand what this problem want's from me (I don't get this "largest" part), I tried to improvise something, so I went like this:
\(\displaystyle x1^3+x2^3\)=
=\(\displaystyle (x1+x2)(x1^2-x1x2+x2^2)=\)
=\(\displaystyle 1[(x1+x2)^2-3x1x2]=\)
=\(\displaystyle 1[1-3(m^2+2m-3)]=\)
=\(\displaystyle 1(1-3m^2-6m+9)=\)
=\(\displaystyle 1(-3m^2-6m+10)\)
Since I have no idea what the problem asks me to do, I just tried to set up an inequality so that \(\displaystyle -3m^2-6m+10\) is less than 0. That leads me to nowhere x)
If you want to help, please do so through the looking glass of Vieta, since this problem is under that section.
I'm really stuck with one problem. Here it goes:
1) \(\displaystyle x1\) and \(\displaystyle x2\) are solutions of the equation \(\displaystyle x^2-x+m^2+2m-3=0\). Value of the real parameter \(\displaystyle m\), for which the sum \(\displaystyle x1^3+x2^3\) is largest, belongs to which set (here they give a bunch of answers, one of which is the correct one).
Since I don't really understand what this problem want's from me (I don't get this "largest" part), I tried to improvise something, so I went like this:
\(\displaystyle x1^3+x2^3\)=
=\(\displaystyle (x1+x2)(x1^2-x1x2+x2^2)=\)
=\(\displaystyle 1[(x1+x2)^2-3x1x2]=\)
=\(\displaystyle 1[1-3(m^2+2m-3)]=\)
=\(\displaystyle 1(1-3m^2-6m+9)=\)
=\(\displaystyle 1(-3m^2-6m+10)\)
Since I have no idea what the problem asks me to do, I just tried to set up an inequality so that \(\displaystyle -3m^2-6m+10\) is less than 0. That leads me to nowhere x)
If you want to help, please do so through the looking glass of Vieta, since this problem is under that section.
Last edited: