Question

Solve each system. Use any method you wish. $$ \left\\{\begin{array}{r} 2 x^{2}+y^{2}=2 \\ x^{2}-2 y^{2}+8=0 \end{array}\right. $$

Short answer

No real solutions.

Step-by-step solution

Simplify the Equations

Start by looking at the given system of equations: \[\begin{equation}\left\{\begin{array}{c}2x^{2} + y^{2} = 2 \ x^{2} - 2 y^{2} + 8 = 0\end{array}\right.\end{equation}\]. Simplify the second equation to isolate the terms involving variables: \[\begin{equation}x^{2} - 2 y^{2} = -8\end{equation}\].

Isolate one variable

From the first equation, solve for one variable in terms of the other. We choose to solve for y: \[\begin{equation}y^{2} = 2 - 2x^{2}\end{equation}\]. Substituting this into the second equation, we get: \[\begin{equation}x^{2} - 2(2 - 2x^{2}) = -8\end{equation}\].

Solve for x

Simplify the new equation in terms of x: \[\begin{equation}x^{2} - 4 + 4x^{2} = -8\end{equation}\]. Combine like terms to get \[\begin{equation}5x^{2} - 4 = -8\end{equation}\] which simplifies to \[\begin{equation}5x^{2} = -4\end{equation}\].

Determine if solutions are feasible

Note that \[\begin{equation}5x^{2} = -4\end{equation}\] cannot have any real solution since a square of a real number cannot be negative. Therefore, there are no real solutions for x.

Key concepts

Quadratic Equations

Quadratic equations are mathematical expressions of the form \( ax^2 + bx + c = 0 \). They are called 'quadratic' because they involve terms raised to the second power (\(x^2\)). In the given problem, the system involves two quadratic equations: \(2x^2 + y^2 = 2\) and \(x^2 - 2y^2 + 8 = 0\). Quadratic equations can have different kinds of solutions: two real solutions, one real solution (a repeated root), or no real solutions. Solving these equations often involves factoring, using the quadratic formula, or, as we'll discuss later, algebraic manipulation.

Substitution Method

The substitution method is a technique used to solve systems of equations by expressing one variable in terms of the other. In our problem, we first simplified the second equation to isolate one variable. By solving \(x^2 - 2y^2 + 8 = 0\), we obtained \(x^2 - 2y^2 = -8\). Next, we solved the first equation for \(y^2\), getting \(y^2 = 2 - 2x^2\), and substituted this into the second simplified equation. This process allows us to reduce the system into a single-variable equation, making it easier to find the solutions.

Real Number Solutions

Real number solutions are values that satisfy the equations without involving imaginary or complex numbers. In our system, after substitution, we reached the equation \(5x^2 = -4\). To determine if there are real solutions, consider that squaring any real number should yield a non-negative result. Since \(5x^2 = -4\) implies that \(x^2 = -\frac{4}{5}\), we see that \(x^2\) cannot be negative if \(x\) is a real number. Therefore, the system has no real number solutions because the calculation results in a non-real number condition.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations to make solving them more manageable. For example, in our system, we simplified the second equation to \(x^2 - 2y^2 = -8\) and substituted \(y^2\) from the first equation. We then combined like terms and simplified to \(5x^2 = -4\). Combining similar terms and simplifying expressions makes it easier to find solutions or determine if solutions exist. It's a critical step in solving complex systems of equations and requires practice to execute smoothly.