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Find the number of real solutions to the system y = x^2 - 5, x^2 + y^2 = 25.

 Jun 11, 2020
 #1
avatar+1128 
+1

so we get y=4 and  x=3since the only integer for x^2+y^2=25 is 3,4,5 since \(\sqrt{25} \)=5 and by the pythagorean theorem either x or y=3 or 4 but to be specific we must check and ony 4=3^2-5 works so x=3 , y=4

 Jun 11, 2020
 #2
avatar+8340 
0

We can actually solve the equation to see how many real solutions there are.

 

\(\begin{cases}y = x^2 - 5\\x^2 + y^2 = 25\end{cases}\\ x^2 + (x^2 - 5)^2 = 25\\ x^4 - 9x^2 = 0\\ x^2 = 0 \text{ or } x^2 = 9\\ x = -3, 0, 3\)

 

Plugging in these values into the equation gives

\(y = 4, -5, 4\)

in this order.

 

Therefore there are 3 pairs of real solution, namely \((-3, 4), (0, -5), \text{ and }(3, 4)\)

 Jun 11, 2020

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