To solve a quadratic equation of the form \( ax^2 + bx + c = 0 \), you can use several methods. The most common methods are factoring, completing the square, and using the quadratic formula. Here’s a step-by-step guide for each method:
1. Factoring
If the quadratic can be factored, you can express it as: \[ (a x + m)(b x + n) = 0 \] where \( m \) and \( n \) are numbers that satisfy:
- \( m \cdot n = c \) (the constant term)
- \( a \cdot n + b \cdot m = b \) (the middle coefficient)
Once factored, set each factor to zero: \[ a x + m = 0 \quad \text{or} \quad b x + n = 0 \] Solve for \( x \) in each case.
2. Completing the Square
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Move the constant term \( c \) to the other side of the equation: \[ ax^2 + bx = -c \]
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Divide the entire equation by \( a \) (if \( a \neq 0 \)): \[ x^2 + \frac{b}{a}x = -\frac{c}{a} \]
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Take half of the coefficient of \( x \) (which is \( \frac{b}{2a} \)), square it, and add it to both sides: \[ x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = -\frac{c}{a} + \left(\frac{b}{2a}\right)^2 \]
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Rewrite the left side as a squared term: \[ \left(x + \frac{b}{2a}\right)^2 = \text{(right side)} \]
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Take the square root of both sides and solve for \( x \): \[ x + \frac{b}{2a} = \pm \sqrt{\text{(right side)}} \] \[ x = -\frac{b}{2a} \pm \sqrt{\text{(right side)}} \]
3. Quadratic Formula
If the equation cannot be easily factored, or you prefer a formulaic approach, you can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
- Identify the coefficients \( a \), \( b \), and \( c \).
- Calculate the discriminant \( D = b^2 - 4ac \).
- If \( D > 0 \), there are two distinct real roots.
- If \( D = 0 \), there is exactly one real root (a repeated root).
- If \( D < 0 \), there are two complex roots.
The solutions are then given by substituting \( a \), \( b \), and \( c \) into the quadratic formula.
Example
Solve \( 2x^2 + 3x - 2 = 0 \) using the quadratic formula:
- Here, \( a = 2 \), \( b = 3 \), \( c = -2 \).
- The discriminant: \[ D = 3^2 - 4 \cdot 2 \cdot (-2) = 9 + 16 = 25 \]
- Since \( D > 0 \), use the quadratic formula: \[ x = \frac{-3 \pm \sqrt{25}}{2 \cdot 2} = \frac{-3 \pm 5}{4} \] This gives two solutions: \[ x = \frac{2}{4} = \frac{1}{2} \quad \text{and} \quad x = \frac{-8}{4} = -2 \]
So the roots of \( 2x^2 + 3x - 2 = 0 \) are \( x = \frac{1}{2} \) and \( x = -2 \).