Solve the system by the substitution method.
-$\begin{array} { l }
x ^ { 2 } + y ^ { 2 } = 36 \
y = x ^ { 2 } - 6
\end{array}$
A) $( 0,6 ) , ( - \sqrt { 11 } , 5 )$
B) $( 0 , - 6 ) , ( \sqrt { 11 } , 5 )$
C) $( - \sqrt { 11 } , 5 ) , ( 0 , - 6 ) , ( \sqrt { 11 } , 5 )$
D) $( 0,6 ) , ( \sqrt { 11 } , 5 )$

Question

Solve the system by the substitution method.
-$\begin{array} { l } 
x ^ { 2 } + y ^ { 2 } = 36 \
y = x ^ { 2 } - 6
\end{array}$
A) $( 0,6 ) , ( - \sqrt { 11 } , 5 )$
B) $( 0 , - 6 ) , ( \sqrt { 11 } , 5 )$
C) $( - \sqrt { 11 } , 5 ) , ( 0 , - 6 ) , ( \sqrt { 11 } , 5 )$
D) $( 0,6 ) , ( \sqrt { 11 } , 5 )$

Quizplus · Accepted Answer

From the second equation, we can substitute y in the first equation and get an equation in terms of x only: 
$x^2 + (x^2-6)^2 = 36$
Simplifying this, we get 
$2x^4 - 12x^2 + 27 = 0$
Factoring out a 2, we get 
$2(x^2-3)^2 + 3 = 0$
But since $(x^2-3)^2$ is non-negative, this equation has no real solutions. Therefore, the system has no real solutions. However, if we were to consider complex solutions, the equation $2(x^2-3)^2 + 3 = 0$ has complex solutions $x=\pm\sqrt{3\pm i\sqrt{2}}$. Substituting these into $y=x^2-6$ gives us the complex solutions: $(\pm\sqrt{3\pm i\sqrt{2}}, 3\mp 2i\sqrt{3\pm i\sqrt{2}})$. The three real solutions in the answer choices are all incorrect.

Solve the System by the Substitution Method x2+y2=36y=x2−6\begin{array} { l } x ^ { 2 } + y ^ { 2 } = 36 \\y = x ^ { 2 } - 6\end{array}x2+y2=36y=x2−6​

Solve the System by the Substitution Method $\begin{array} { l } x ^ { 2 } + y ^ { 2 } = 36 \\y = x ^ { 2 } - 6\end{array}$