Solving a System of Quadratic Equations: 2x^2 + y^2 = 21 and x^2

Math Problem Statement

Solve the system of equations: 2x^2 + y^2 = 21 and x^2 - 2y^2 = 41. Find the intersection points.

To solve the given system of equations for the intersection points, we are working with:

1. & \; 2x^2 + y^2 = 21, \\ 2. & \; x^2 - 2y^2 = 41. \end{aligned}$$ ### Step 1: Express $$x^2$$ or $$y^2$$ from one equation. Let us express $$x^2$$ from the second equation: $$x^2 = 2y^2 + 41.$$ ### Step 2: Substitute $$x^2$$ into the first equation. Replace $$x^2$$ in the first equation $$2x^2 + y^2 = 21$$: $$2(2y^2 + 41) + y^2 = 21.$$ Simplify: $$4y^2 + 82 + y^2 = 21,$$ $$5y^2 + 82 = 21.$$ $$5y^2 = -61.$$ $$y^2 = -\frac{61}{5}.$$ ### Step 3: Analyze the result. The equation $$y^2 = -\frac{61}{5}$$ is invalid because $$y^2$$ cannot be negative. This implies that there are **no real solutions** to the system of equations. ### Final Answer: The system has no real intersection points. Thus, the solution is: $$\text{DNE (Does Not Exist)}.$$ --- Let me know if you'd like a breakdown of any step! Here are 5 related questions to expand your understanding: 1. How do we handle systems of equations with imaginary solutions? 2. What is the geometric meaning of no intersection between two curves? 3. How can graphing the equations visually confirm that the curves do not intersect? 4. What are alternative methods to solve systems of quadratic equations? 5. Can substitution always simplify non-linear systems like this? **Tip**: Always verify whether the solutions satisfy the domain restrictions (like $$x^2, y^2 \geq 0$$) when solving quadratic equations.

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Algebra
Quadratic Equations
System of Equations

Substitution method
Quadratic equation properties (x^2, y^2)

No real solutions for negative squared terms

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