Question

How do I solve a quadratic equation of the form ax^2 + bx + c = 0?

Answer 1

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
1. Move the constant term \( c \) to the other side of the equation:
\[
ax^2 + bx = -c
\]

2. Divide the entire equation by \( a \) (if \( a \neq 0 \)):
\[
x^2 + \frac{b}{a}x = -\frac{c}{a}
\]

3. Take half of the coefficient of \( x \) (which is \( \frac{b}{2a} \)), square it, and add it to both sides:
\[
x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = -\frac{c}{a} + \left(\frac{b}{2a}\right)^2
\]

4. Rewrite the left side as a squared term:
\[
\left(x + \frac{b}{2a}\right)^2 = \text{(right side)}
\]

5. 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:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
1. Identify the coefficients \( a \), \( b \), and \( c \).
2. Calculate the discriminant \( D = b^2 - 4ac \).
3. If \( D > 0 \), there are two distinct real roots.
4. If \( D = 0 \), there is exactly one real root (a repeated root).
5. 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:
1. Here, \( a = 2 \), \( b = 3 \), \( c = -2 \).
2. The discriminant:
\[
D = 3^2 - 4 \cdot 2 \cdot (-2) = 9 + 16 = 25
\]
3. 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 \( x = -2 \).