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I need a help please

X^2+Y^2=29

Y=X^2+1

Guest Jun 4, 2015

Best Answer 

 #1
avatar+92781 
+10

$$\\x^2+y^2=29\qquad (1)\\
y=x^2+1\\
x^2=y-1 \qquad (2)\\
$Note: y-1 MUST be $ \ge 0 \;\;SO\;\; y\ge 1\\
$sub (2) into (1)$\\
y-1+y^2=29\\
y^2+y-30=0\\
(y-5)(y+6)=0\\
y=5 \;\;or\;\;y=-6\\\\
$y must be greater or equal to 1 so y=-6 is invalid in the real number system$
y=5\;\;sub\;to\;find\;x\\
x^2=5-1=4\\
x=\pm2\\\\
$So the 2 solutions are (2,5)\;\;and\;\;(-2,5)$$

 

Here is a really neat graph to show you what is happening 

 

https://www.desmos.com/calculator/kzutxgb3jm

 

 

Melody  Jun 4, 2015
 #1
avatar+92781 
+10
Best Answer

$$\\x^2+y^2=29\qquad (1)\\
y=x^2+1\\
x^2=y-1 \qquad (2)\\
$Note: y-1 MUST be $ \ge 0 \;\;SO\;\; y\ge 1\\
$sub (2) into (1)$\\
y-1+y^2=29\\
y^2+y-30=0\\
(y-5)(y+6)=0\\
y=5 \;\;or\;\;y=-6\\\\
$y must be greater or equal to 1 so y=-6 is invalid in the real number system$
y=5\;\;sub\;to\;find\;x\\
x^2=5-1=4\\
x=\pm2\\\\
$So the 2 solutions are (2,5)\;\;and\;\;(-2,5)$$

 

Here is a really neat graph to show you what is happening 

 

https://www.desmos.com/calculator/kzutxgb3jm

 

 

Melody  Jun 4, 2015
 #2
avatar+87301 
0

Nice graph, Melody....!!!

 

 

CPhill  Jun 5, 2015
 #3
avatar+92781 
0

Thanks Chris. :)

Melody  Jun 5, 2015

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