There are how many solutions of the system?
$\begin{aligned}
x^{2}+y^{2} &=25 \
y &=x^{2}-7
\end{aligned}$
A) None
B) One
C) Two
D) Three
E) Four

Question

Quizplus · Accepted Answer

Substituting $y = x^2 - 7$ into the first equation, we get $x^2 + (x^2 - 7)^2 = 25$. Simplifying, $x^2 + x^4 - 14x^2 + 49 = 25$, which simplifies to $x^4 - 13x^2 + 24 = 0$. Factoring, we get $(x^2 - 4)(x^2 - 9) = 0$, which gives $x^2 = 4$ or $x^2 = 9$, leading to $x = \pm2$ or $x = \pm3$. For each value of $x$, there is a corresponding $y$ value, resulting in four solutions.

There Are How Many Solutions of the System x2+y2=25y=x2−7\begin{aligned}x^{2}+y^{2} &=25 \\y &=x^{2}-7\end{aligned}x2+y2y​=25=x2−7​

There Are How Many Solutions of the System $\begin{aligned}x^{2}+y^{2} &=25 \\y &=x^{2}-7\end{aligned}$