[SOLVED] Find number of distinct pairs (x,y)?(difficult)

• October 11th 2008, 03:39 AM
fardeen_gen
[SOLVED] Find number of distinct pairs (x,y)?(difficult)
The number of distinct pairs (x,y) of real numbers satisfying x = x^3 + y^4 and y = 2xy is:

A)5
B)12
C)3
D)7
• October 11th 2008, 04:39 AM
skeeter
$y = 2xy$

$y - 2xy = 0$

$y(1 - 2x) = 0$

$y = 0$ or $x = \frac{1}{2}$

for $y = 0$ ...

$x = x^3 + y^4$

$x = x^3$

$x - x^3 = 0$

$x(1 - x^2) = x(1 - x)(1 + x) = 0$

$(0,0)$, $(1,0)$ and $(-1,0)$

for $x = \frac{1}{2}$ ...

$\frac{1}{2} = \frac{1}{8} + y^4$

$y^4 - \frac{3}{8} = 0$

$\left(y^2 + \sqrt{\frac{3}{8}}\right)\left(y^2 - \sqrt{\frac{3}{8}}\right) = 0$

$\left(y^2 + \sqrt{\frac{3}{8}}\right)\left(y - \sqrt[4]{\frac{3}{8}}\right)\left(y + \sqrt[4]{\frac{3}{8}}\right) = 0$

$\left(\frac{1}{2}, \sqrt[4]{\frac{3}{8}}\right)$ and $\left(\frac{1}{2}, -\sqrt[4]{\frac{3}{8}}\right)$

looks like 5 distinct solutions to me ... hope I didn't miss any.
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