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Chapter 12: Problem 26

Solve each system using the substitution method. \(x^{2}+y^{2}=4\) \(y=x^{2}-2\)

Short Answer

Expert verified
The solutions are (0, -2), (\sqrt{3}, 1), and (-\sqrt{3}, 1).

Step by step solution

01

- Substitute the second equation into the first

The second equation is given by: \( y = x^2 - 2 \) Substituting this into the first equation: \( x^2 + (x^2 - 2)^2 = 4 \)
02

- Expand and simplify the equation

Simplify the equation by expanding \( (x^2 - 2)^2 \): \[ (x^2 - 2)^2 = (x^2 - 2)(x^2 - 2) = x^4 - 4x^2 + 4 \] So, the equation becomes: \[ x^2 + x^4 - 4x^2 + 4 = 4 \]
03

- Combine like terms and solve for x

Combine the like terms: \[ x^4 - 3x^2 + 4 = 4 \] Subtract 4 from both sides: \[ x^4 - 3x^2 = 0 \] Factor out \( x^2 \): \[ x^2(x^2 - 3) = 0 \] So, \( x^2 = 0 \) or \( x^2 = 3 \). Then \( x = 0 \) or \( x = \pm \sqrt{3} \)
04

- Solve for y when x = 0

When \( x = 0 \): \[ y = x^2 - 2 = 0^2 - 2 = -2 \]
05

- Solve for y when x = \pm\sqrt{3}

When \( x = \sqrt{3} \): \[ y = (\sqrt{3})^2 - 2 = 3 - 2 = 1 \] When \( x = -\sqrt{3} \): \[ y = (-\sqrt{3})^2 - 2 = 3 - 2 = 1 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Systems of Equations
Systems of equations are sets of two or more equations that have common solutions. To solve these systems, we can use various methods like graphing, substitution, or elimination.

When dealing with systems of equations involving quadratic equations, we can use the substitution method. This method involves replacing one variable with an equivalent expression from another equation.
Quadratic Equations
A quadratic equation is a type of polynomial equation of degree 2. It is usually written in the form: \( ax^2 + bx + c = 0 \).

In the given exercise, we have a quadratic equation: \( y = x^2 - 2 \). Our goal is to substitute this equation into another to find the solutions of the system.
Solving Algebraic Equations
Solving algebraic equations involves finding the values of the variables that make the equation true. Here, we start by substituting the second equation into the first:
  • Given: \(x^{2}+y^{2}=4\)
  • Substitute \( y = x^2 - 2 \): \( x^2 + (x^2 - 2)^2 = 4 \)

Next, we expand \( (x^2 - 2)^2 \): \[ (x^2 - 2)^2 = x^2 - 4x^2 + 4 \rightarrow x^4 - 4x^2 + 4 \].
Substituting back, we get: \( x^2 + x^4 - 4x^2 + 4 = 4 \).
Combining like terms simplifies this to: \( x^4 - 3x^2 = 0 \).
Factoring
Factoring is a method used to break down algebraic expressions into simpler parts. In our case, we factor out \( x^2 \) from the equation: \( x^4 - 3x^2 = 0 \), resulting in:
\[ x^2 (x^2 - 3) = 0 \]

This gives us two possible equations: \( x^2 = 0 \) and \( x^2 = 3 \). Solving these, we get: \( x = 0 \) or \( x = \pm \sqrt{3} \).
  • When \( x = 0 \): \( y = x^2 - 2 = -2 \).
  • When \( x = \sqrt{3} \): \( y = 1 \).
  • When \( x = -\sqrt{3} \): \( y = 1 \).

Therefore, the solutions are: \( (0, -2) \), \( ( \sqrt{3}, 1) \), and \( ( -\sqrt{3}, 1) \).

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