Determine the real solutions to the system of nonlinear equations.
-$$\begin{array} { l }
x ^ { 3 } + y = 0 \
6 x ^ { 2 } - y = 0
\end{array}$$
A) $( 0,0 ) , ( - 6,36 )$
B) $( - 1,1 ) , ( - 6,216 )$
C) $( 0,0 ) , ( 6 , - 216 )$
D) $( 0,0 ) , ( - 6,216 )$

Question

Determine the real solutions to the system of nonlinear equations.
-$$\begin{array} { l } 
x ^ { 3 } + y = 0 \
6 x ^ { 2 } - y = 0
\end{array}$$
A) $( 0,0 ) , ( - 6,36 )$
B) $( - 1,1 ) , ( - 6,216 )$
C) $( 0,0 ) , ( 6 , - 216 )$
D) $( 0,0 ) , ( - 6,216 )$

Quizplus · Accepted Answer

By adding the two equations, we get $x^3 + 6x^2 = 0$, which factors to $x^2(x + 6) = 0$. This gives $x = 0$ or $x = -6$. Substituting $x = 0$ into the original equations gives $y = 0$, and substituting $x = -6$ gives $y = 216$. Thus, the solutions are $(0,0)$ and $(-6,216)$.

Determine the Real Solutions to the System of Nonlinear Equations x3+y=06x2−y=0\begin{array} { l } x ^ { 3 } + y = 0 \\6 x ^ { 2 } - y = 0\end{array}x3+y=06x2−y=0​

Determine the Real Solutions to the System of Nonlinear Equations $\begin{array} { l } x ^ { 3 } + y = 0 \\6 x ^ { 2 } - y = 0\end{array}$