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Solve the system x^2 = y + 2, y^2 = x + 2.

 Dec 1, 2019
 #1
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Do plugin

x = y^2-2

x^2 = (y^2-2)(y^2-2) = y^4-4y^2+4. So y^4-4y^2+4=y+2. Then y = y^4-4y^2+2. Begin to test values starting with 5 and getting smaller. For 5, it doesn't work. It is very far, so we try 3. For 3, we have 81 = 36+2 which is false. For 2, we have 2 = 16-16+2 which is true so y = 2 and x = 2 clearly.

 Dec 1, 2019
 #2
avatar+24401 
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Solve the system \(x^2 = y + 2,\ y^2 = x + 2\).

 

\(\begin{array}{|lrcll|} \hline (1) & x^2 &=& y+2 \quad \text{or} \quad \boxed{y = x^2-2} \\\\ (2) & y^2 &=& x+2 \\ & \left(x^2-2\right)^2 &=& x+2 \\ & x^4-4x^2+4 &=& x+2 \\ & x^4-4x^2-x+2 &=& 0 \\\\ \boxed{x_1= 2}: & x^4-4x^2-x+2 &\overset{?}=& 0 \\ & 2^4-4*2^2-2+2 &=& 0 \ \checkmark \\ & \text{Polynomial long division}: & \\ & (x^4-4x^2-x+2) &=& (x-2)(x^3+2x^2-1) \\\\ \boxed{x_2= -1}: & x^3+2x^2-1 &\overset{?}=& 0 \\ & (-1)^3+2(-1)^2-1 &=& 0 \ \checkmark \\ & \text{Polynomial long division}: & \\ & (x^3+2x^2-1) &=& (x+1)(x^2+x-1) \\\\ & x^2+x-1 &=& 0 \\\\ & x &=& \dfrac{-1\pm \sqrt{1-4(-1)}} {2} \\ & x &=& \dfrac{-1\pm \sqrt{5}} {2} \\ \boxed{x_3 = \dfrac{-1+\sqrt{5}} {2}} \\ \boxed{x_4 = \dfrac{-1-\sqrt{5}} {2}} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline && y = x^2-2 \\\\ x_1=1 && y_1 = -1 \\ x_2 =2 && y_2 = 2 \\ x_3 = \dfrac{-1+\sqrt{5}} {2} && y_3 = \dfrac{-1-\sqrt{5}} {2} \\ x_4 = \dfrac{-1-\sqrt{5}} {2} && y_4 = \dfrac{-1+\sqrt{5}} {2} \\ \hline \end{array} \)

 

laugh

 Dec 2, 2019

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