Answer
[tex]\begin{gathered} (\sqrt[]{6},0) \\ (-\sqrt[]{6},0) \\ (\sqrt[]{5},-1) \\ (-\sqrt[]{5},-1) \end{gathered}[/tex]
Explanation
Given quadratic-quadratic system of equation:
[tex]\begin{gathered} x^2+y^2=6----i \\ x^2-y=6----ii \end{gathered}[/tex]
From (i):
[tex]x^2=6-y^2----iii[/tex]
From (ii) also:
[tex]x^2=6+y----iv[/tex]
(iii) = (iv) implies:
[tex]\begin{gathered} 6-y^2=6+y \\ Combine\text{ the like terms} \\ y^2+y=6-6 \\ y^2+y=0 \\ By\text{ factorization} \\ y(y+1)=0 \\ \text{Either }y=0\text{ or }y+1=0 \\ y=0\text{ or }y=-1 \end{gathered}[/tex]
To solve for the values of x, substitute y = 0 and y = -1 into (iv):
[tex]\begin{gathered} \text{Recall (iv)} \\ x^2=6+y \\ \text{For }y=0 \\ x^2=6+0 \\ x^2=6 \\ \text{Take the square root of both sides} \\ x=\pm\sqrt[]{6} \\ \therefore when\text{ }y=0,\text{ }x=\pm\sqrt[]{6}\text{ } \\ \text{Hence, }(\pm\sqrt[]{6},0) \\ \\ \text{For For }y=-1 \\ x^2=6+(-1) \\ x^2=5 \\ \text{Take square root of both sides} \\ x=\pm\sqrt[]{5} \\ \therefore when\text{ }y=-1,\text{ }x=\pm\sqrt[]{5} \\ Hence,\text{ }(\pm\sqrt[]{5},-1) \end{gathered}[/tex]
Therefore the solution of the system of equations are:
[tex]\begin{gathered} (\sqrt[]{6},0) \\ (-\sqrt[]{6},0) \\ (\sqrt[]{5},-1) \\ (-\sqrt[]{5},-1) \end{gathered}[/tex]
