Certainly! Here’s a quick guide on how to determine the solution sets of quadratic equations.
Quadratic Equation Format
A quadratic equation can be expressed in standard form as: \[ ax^2 + bx + c = 0 \] where \( a \), \( b \), and \( c \) are constants and \( a \neq 0 \).
Methods to Solve Quadratic Equations
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Factoring
- If the quadratic can be factored, write it as: \[ (px + q)(rx + s) = 0 \]
- Set each factor to zero: \[ px + q = 0 \quad \text{and} \quad rx + s = 0 \]
- Solve for \( x \).
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Quadratic Formula
- If the quadratic is not easily factorable, use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
- Here, \( b^2 - 4ac \) is called the discriminant.
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Completing the Square
- Rearrange the equation to isolate the quadratic and linear terms: \[ ax^2 + bx = -c \]
- Divide by \( a \): \[ x^2 + \frac{b}{a}x = -\frac{c}{a} \]
- Complete the square on the left-hand side and solve for \( x \).
Analyzing the Discriminant
The discriminant \( D = b^2 - 4ac \) determines the nature of the roots:
- If \( D > 0 \): Two distinct real solutions.
- If \( D = 0 \): One real double solution (or two real solutions that are the same).
- If \( D < 0 \): Two complex (imaginary) solutions.
Example Problems
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Example 1: Solve \( x^2 - 5x + 6 = 0 \).
- Factor: \( (x - 2)(x - 3) = 0 \)
- Solutions: \( x = 2 \) and \( x = 3 \).
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Example 2: Solve \( 2x^2 + 4x + 2 = 0 \).
- Discriminant: \( D = 4^2 - 4\cdot2\cdot2 = 16 - 16 = 0 \).
- Solution: \( x = \frac{-4}{2\cdot2} = -1 \) (double root).
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Example 3: Solve \( x^2 + 2x + 5 = 0 \).
- Discriminant: \( D = 2^2 - 4\cdot1\cdot5 = 4 - 20 = -16 \).
- Solutions: \( x = \frac{-2 \pm \sqrt{-16}}{2\cdot1} = -1 \pm 2i \) (complex roots).
Quick Check Questions
- Solve \( x^2 - 4 = 0 \).
- Solve \( 3x^2 + 12x + 12 = 0 \).
- Determine the nature of the roots for \( x^2 + 6x + 10 = 0 \).
Feel free to ask for specific examples or further clarification on any of the methods!
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