Originally Posted by
nathan02079 1. $\displaystyle \frac{3}{2}y^2-2=y$ Mr F says: $\displaystyle \therefore 3y^2 - 2y - 4 = 0$, solved in several ways eg. by using the quadratic formula.
2. $\displaystyle \frac{2}{y}=\frac{3}{y^2}+2$ Mr F says: $\displaystyle \therefore 2y = 3 + 2y^2 \therefore 2y^2 - 2y + 3 = 0$, which has no real solutions.
3. $\displaystyle \sqrt3x^2=8\sqrt2x-4\sqrt3$ Mr F says: $\displaystyle \therefore 3x^2 = 8\sqrt{6}x - 12 \therefore 3x^2 - 8\sqrt{6} x + 12 = 0$, solved in several ways eg. by using the quadratic formula.
Thanks...those three just trouble me =P