• Jan 13th 2008, 09:29 PM
nathan02079
1. $\frac{3}{2}y^2-2=y$
2. $\frac{2}{y}=\frac{3}{y^2}+2$
3. $\sqrt3x^2=8\sqrt2x-4\sqrt3$

Thanks...those three just trouble me =P
• Jan 13th 2008, 09:43 PM
mr fantastic
Quote:

Originally Posted by nathan02079
1. $\frac{3}{2}y^2-2=y$ Mr F says: $\therefore 3y^2 - 2y - 4 = 0$, solved in several ways eg. by using the quadratic formula.

2. $\frac{2}{y}=\frac{3}{y^2}+2$ Mr F says: $\therefore 2y = 3 + 2y^2 \therefore 2y^2 - 2y + 3 = 0$, which has no real solutions.

3. $\sqrt3x^2=8\sqrt2x-4\sqrt3$ Mr F says: $\therefore 3x^2 = 8\sqrt{6}x - 12 \therefore 3x^2 - 8\sqrt{6} x + 12 = 0$, solved in several ways eg. by using the quadratic formula.

Thanks...those three just trouble me =P

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