+0  
 
0
120
1
avatar

Positive real numbers x, y  satisfy the equations x^2 + y^2 = 1 and x^4 + y^4 = 17/19. Find xy.

 May 14, 2021
 #1
avatar+26213 
+1

Positive real numbers \(x\), \(y\)  satisfy the equations \(x^2 + y^2 = 1\) and
\(x^4 + y^4 = \dfrac{17}{19}\).
Find \(xy\).

 

\(\begin{array}{|rcll|} \hline \left( x^2 + y^2\right)^2 &=& x^4 + 2x^2y^2 + y^4 \\ \left( x^2 + y^2\right)^2 &=& x^4 + y^4 + 2x^2y^2 \quad | \quad x^2 + y^2 = 1,~ \dfrac{17}{19} \\\\ 1^2 &=& \dfrac{17}{19} + 2x^2y^2 \\\\ 2x^2y^2 &=& 1 - \dfrac{17}{19} \\\\ 2x^2y^2 &=& \dfrac{2}{19} \quad | \quad :2 \\\\ x^2y^2 &=& \dfrac{1}{19} \quad | \quad \text{sqrt both sides} \\\\ \mathbf{ xy } &=& \mathbf{ \dfrac{1}{\sqrt{19}} } \\ \hline \end{array}\)

 

laugh

 May 14, 2021

4 Online Users

avatar
avatar