Solving quadratic equations using the quadratic formula is a fundamental technique in algebra. A quadratic equation is typically in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are coefficients and \( x \) represents the unknown variable. The quadratic formula is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This formula is derived from completing the square in the general form of a quadratic equation. It provides the solutions for \( x \) by considering all possible scenarios (real and complex solutions).
### Examples:
#### Example 1: Real and Distinct Roots
Consider the equation \( 2x^2 - 4x - 6 = 0 \).
1. Identify \( a = 2 \), \( b = -4 \), and \( c = -6 \).
2. Plug these into the formula:
\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(-6)}}{2(2)} \]
3. Simplify the expression:
\[ x = \frac{4 \pm \sqrt{16 + 48}}{4} \]
\[ x = \frac{4 \pm \sqrt{64}}{4} \]
\[ x = \frac{4 \pm 8}{4} \]
4. Find the two solutions:
\[ x_1 = \frac{4 + 8}{4} = 3 \]
\[ x_2 = \frac{4 - 8}{4} = -1 \]
#### Example 2: Real and Repeated Roots
Consider \( x^2 - 6x + 9 = 0 \).
1. Here, \( a = 1 \), \( b = -6 \), \( c = 9 \).
2. Apply the formula:
\[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(9)}}{2(1)} \]
3. Simplify:
\[ x = \frac{6 \pm \sqrt{36 - 36}}{2} \]
\[ x = \frac{6 \pm 0}{2} \]
4. The single (repeated) solution is:
\[ x = \frac{6}{2} = 3 \]
#### Example 3: Complex Roots
Consider \( x^2 + 4x + 8 = 0 \).
1. Identify \( a = 1 \), \( b = 4 \), \( c = 8 \).
2. Insert into the formula:
\[ x = \frac{-4 \pm \sqrt{4^2 - 4(1)(8)}}{2(1)} \]
3. Simplify:
\[ x = \frac{-4 \pm \sqrt{16 - 32}}{2} \]
\[ x = \frac{-4 \pm \sqrt{-16}}{2} \]
4. Solve for complex solutions:
\[ x_1 = \frac{-4 + 4i}{2} = -2 + 2i \]
\[ x_2 = \frac{-4 - 4i}{2} = -2 - 2i \]
In these examples, the quadratic formula effectively finds the roots of the quadratic equations, whether they are real (distinct or repeated) or complex.
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