Solve: $\left\{ \begin{aligned}
x ^ { 2 } + y ^ { 2 } & = 182 \
y & = x ^ { 2 }
\end{aligned} \right.$
A) $\{ ( \sqrt { 13 } , 13 ) , ( - \sqrt { 13 } , 13 ) \}$
B) $\{ ( \sqrt { 13 } , 13 ) \}$
C) $\{ ( 169,13 ) \}$
D) no solution

Question

Quizplus · Accepted Answer

Substituting $y = x^2$ into the first equation gives $x^2 + (x^2)^2 = 182$, which simplifies to $x^4 + x^2 - 182 = 0$. This can be solved as a quadratic equation in terms of $x^2$, yielding solutions for $x^2 = 13$. Thus, $x = \pm \sqrt{13}$, and $y = 13$ for both cases.

Solve {x2+y2=182y=x2\left\{ \begin{aligned}x ^ { 2 } + y ^ { 2 } & = 182 \\y & = x ^ { 2 }\end{aligned} \right.{x2+y2y​=182=x2​

Solve $\left\{ \begin{aligned}x ^ { 2 } + y ^ { 2 } & = 182 \\y & = x ^ { 2 }\end{aligned} \right.$