Problem 28 - Precalculus Mathematics for Calculus

Question

Find all solutions of the system of equations. $$\left\\{\begin{aligned} x^{2}+2 y^{2} &=2 \\ 2 x^{2}-3 y &=15 \end{aligned}\right.$$

Step-by-Step Solution

Answer

No real solutions exist for the given system of equations.

Step 1: Analyze the System of Equations

We have a system with two equations: \(x^{2} + 2y^{2} = 2\) and \(2x^{2} - 3y = 15\). We need to find all pairs \((x, y)\) that satisfy both equations simultaneously.

Step 2: Solve for \(x^2\) in the First Equation

From the first equation \(x^2 + 2y^2 = 2\), solve for \(x^2\) to get \(x^2 = 2 - 2y^2\).

Step 3: Substitute \(x^2\) in the Second Equation

Replace \(x^2\) in the second equation with the expression from Step 2: \(2(2 - 2y^2) - 3y = 15\). Simplify to get an equation in terms of \(y\) only.

Step 4: Simplify the Equation

Distribute and simplify: \[4 - 4y^2 - 3y = 15\] which leads to \[-4y^2 - 3y - 11 = 0\]. Multiply the entire equation by \(-1\) to get \[4y^2 + 3y + 11 = 0\].

Step 5: Solve the Quadratic Equation for \(y\)

Find \(y\) using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4\), \(b = 3\), and \(c = 11\). Calculate the discriminant \(b^2 - 4ac = 3^2 - 4 \times 4 \times 11 = 9 - 176 = -167\), which is negative.

Step 6: Interpret the Results

A negative discriminant indicates that there are no real solutions for \(y\). Therefore, the system of equations has no real solutions in terms of \(x\) and \(y\).

Key Concepts

Quadratic Equations

Quadratic equations are an essential part of algebra. They often appear in the form \( ax^2 + bx + c = 0 \). In our given system, the quadratic equation describes how values of \( x \) and \( y \) relate through their squares.

In the example exercise, our first equation \( x^2 + 2y^2 = 2 \) is quadratic because it contains terms with squares of variables. Understanding the standard form of a quadratic equation helps in rearranging and solving these equations.

Quadratic equations can be visualized as parabolas when graphed on a coordinate plane. The solutions are the x-values where the parabola intersects the x-axis. In systems of equations, we look for points of intersection between curves represented by these equations.

Recognizing when an equation can be rewritten in a quadratic form, as we did in the outlined steps, is crucial in solving these problems, especially when they are part of a system of equations.

Substitution Method

The substitution method is a strategic procedure to solve systems of equations. It involves expressing one variable in terms of another using one equation and substituting this expression into the other equation.

In the exercise, we first rearrange the first equation to solve for \( x^2 \):

• The expression becomes \( x^2 = 2 - 2y^2 \).

This expression allows us to substitute \( x^2 \) in the second equation:

• This step simplifies a system with two unknowns into a single equation in terms of \( y \).

Substitution is a powerful method because it reduces complexity, enabling us to solve the problem step by step. It's especially useful when dealing with non-linear equations, where solutions aren't immediately obvious. By simplifying one equation, we effectively narrow down the possible solutions, making the problem more manageable.

Real Solutions

When solving systems of equations, real solutions refer to solutions that exist on the coordinate plane, meaning they involve real numbers rather than imaginary numbers.

In our example, when solving for \( y \), the discriminant from the quadratic formula \( b^2 - 4ac \) is negative:

• A negative discriminant indicates no intersection points on a real plane, leading to no real solutions.

It's important to calculate the discriminant to understand the nature of solutions. Here, the calculation showed \( -167 \), implying that the solutions for \( y \) lie within the complex plane and cannot be graphed on the traditional x-y coordinate system.

Recognizing when there are no real solutions is significant, as it affects how we interpret and conclude our findings in applied mathematics problems.