Question

How to solve quadratic equations

Answer 1

Quadratic equations can be solved using several methods. A quadratic equation generally takes the form:

\[ ax^2 + bx + c = 0 \]

where \( a \), \( b \), and \( c \) are constants, and \( a \neq 0 \). Here are the common methods to solve such equations:

### 1. Factoring

If the quadratic can be factored, you can write it as:

\[ (px + q)(rx + s) = 0 \]

To find the solutions, set each factor equal to zero:

\[ px + q = 0 \quad \text{and} \quad rx + s = 0 \]

Solve for \( x \) in both equations.

**Example**:

Solve \( x^2 - 5x + 6 = 0 \).

Factoring gives:

\[ (x - 2)(x - 3) = 0 \]

So, \( x - 2 = 0 \) or \( x - 3 = 0 \) implies \( x = 2 \) or \( x = 3 \).

### 2. Using the Quadratic Formula

If factoring is not straightforward, you can use the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

This formula will give you the solutions regardless of whether the quadratic can be factored.

**Example**:

Solve \( 2x^2 - 4x - 6 = 0 \).

Here, \( a = 2 \), \( b = -4 \), and \( c = -6 \). First, calculate the discriminant:

\[ b^2 - 4ac = (-4)^2 - 4(2)(-6) = 16 + 48 = 64 \]

Now, apply the quadratic formula:

\[ x = \frac{-(-4) \pm \sqrt{64}}{2(2)} = \frac{4 \pm 8}{4} \]

This gives you two solutions:

1. \( x = \frac{12}{4} = 3 \)
2. \( x = \frac{-4}{4} = -1 \)

### 3. Completing the Square

This method involves rewriting the quadratic in the form:

\[ (x - p)^2 = q \]

Then, solve for \( x \) by taking the square root.

**Example**:

Solve \( x^2 - 4x - 5 = 0 \).

First, rearrange it:

\[ x^2 - 4x = 5 \]

Now, complete the square:

1. Take half of the coefficient of \( x \), square it: \( (-4/2)^2 = 4 \)
2. Add this to both sides:

\[ x^2 - 4x + 4 = 9 \]

This gives:

\[ (x - 2)^2 = 9 \]

Now, take the square root:

\[ x - 2 = \pm 3 \]

So, \( x = 5 \) or \( x = -1 \).

### Summary

You can choose the method that best suits the situation, though the quadratic formula is a reliable option for any quadratic equation.