+0

# I need a help please

0
523
3

X^2+Y^2=29

Y=X^2+1

Jun 4, 2015

#1
+95361
+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

Jun 4, 2015

#1
+95361
+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
#2
+94618
0

Nice graph, Melody....!!!

Jun 5, 2015
#3
+95361
0

Thanks Chris. :)

Jun 5, 2015