Two Quadratics Equations

**DivideBy0** · July 26th 2007, 01:31 AM

Find all possible values for x and y given the equations $x+y^2 = 2$ and $x^2 + y = 2$ .

**Plato** · July 26th 2007, 03:54 AM

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

**Soroban** · July 26th 2007, 03:58 AM

Hello, DivideBy0!

Find all possible values for x and y given: . $\begin{array}{cc}x+y^2 \:= \:2 & [1]\\ x^2 + y \:= \:2& [2]\end{array}$

From [2], we have: . $y\:=\:2-x^2$

Substitute into [1]: . $x + (2-x^2)^2 \;=\;2\quad\Rightarrow\quad x + 4 - 4x^2 + x^4\:=\:2$

We have the quartic: . $x^4 - 4x^2 + x + 2 \:=\:0$

. . which factors: . $(x - 1)(x + 2)(x^2 - x - 1)\:=\:0$

. . and has roots: . $x \;=\;1,\:\text{-}2,\:\frac{1\pm\sqrt{5}}{2}$

The corresponding $y$ -values are: . $y \;=\;1,\:\text{-}2,\:\frac{1\mp\sqrt{5}}{2}$

Therefore: . $(x,\,y) \;\;=\;\;(1,\,1),\;(\text{-}2,\,\text{-}2),\;\left(\frac{1+\sqrt{5}}{2},\,\frac{1-\sqrt{5}}{2}\right),\;\left(\frac{1-\sqrt{5}}{2},\,\frac{1+\sqrt{5}}{2}\right)$

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