Question

Solving a Quadratic-Quadratic SvatOfChoose the solution(s) of the following system of equations:x² + y² = 6x2 - y = 6+no solution(6.0)(5.1)(5.-1)

Answer 1

`\begin{gathered} (\sqrt[]{6},0) \\ (-\sqrt[]{6},0) \\ (\sqrt[]{5},-1) \\ (-\sqrt[]{5},-1) \end{gathered}`

Step-by-step explanation

Given quadratic-quadratic system of equation:

`\begin{gathered} x^2+y^2=6----i \\ x^2-y=6----ii \end{gathered}`

From (i):

`x^2=6-y^2----iii`

From (ii) also:

`x^2=6+y----iv`

(iii) = (iv) implies:

`\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}`

To solve for the values of x, substitute y = 0 and y = -1 into (iv):

`\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}`

Therefore the solution of the system of equations are:

`\begin{gathered} (\sqrt[]{6},0) \\ (-\sqrt[]{6},0) \\ (\sqrt[]{5},-1) \\ (-\sqrt[]{5},-1) \end{gathered}`