How to Solve Any Quadratic Equation Using the “Almighty” Formula

Last updated on July 30th, 2024 at 09:22 pm

How to Solve Any Quadratic Equation Using the “Almighty” Formula

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Quadratic equations are polynomial equations of the form:

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

where \( a \), \( b \), and \( c \) are constants, with \( a \neq 0 \). The solutions to these equations can be found using the quadratic formula, which is:

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

This formula gives the roots of the quadratic equation, which are the values of \( x \) that satisfy the equation. Let’s break down the steps to solve a quadratic equation using the quadratic formula.

Step-by-Step Guide

Step 1: Identify the coefficients
The first step is to identify the coefficients \( a \), \( b \), and \( c \) in the quadratic equation. For example, consider the equation:

\[ 2x^2 – 4x – 6 = 0 \]

Here, \( a = 2 \), \( b = -4 \), and \( c = -6 \).

Step 2: Calculate the discriminant
The discriminant (\( \Delta \)) is the part of the quadratic formula under the square root. It is calculated as:

\[ \Delta = b^2 – 4ac \]

Using our example:

\[ \Delta = (-4)^2 – 4 \cdot 2 \cdot (-6) \]
\[ \Delta = 16 + 48 \]
\[ \Delta = 64 \]

The discriminant helps determine the nature of the roots:
– If \( \Delta > 0 \), there are two distinct real roots.
– If \( \Delta = 0 \), there is one real root (a repeated root).
– If \( \Delta < 0 \), there are two complex roots.Step 3: Apply the quadratic formula With the discriminant calculated, substitute \( a \), \( b \), and \( \Delta \) back into the quadratic formula:\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \]For our example:\[ x = \frac{-(-4) \pm \sqrt{64}}{2 \cdot 2} \] \[ x = \frac{4 \pm 8}{4} \]Step 4: Simplify the solutions Solve for the two possible values of \( x \):\[ x_1 = \frac{4 + 8}{4} = \frac{12}{4} = 3 \]\[ x_2 = \frac{4 - 8}{4} = \frac{-4}{4} = -1 \]So, the solutions to the equation \( 2x^2 - 4x - 6 = 0 \) are \( x = 3 \) and \( x = -1 \).Example 1: Positive Discriminant Solve the quadratic equation \( x^2 - 3x + 2 = 0 \).1. Identify the coefficients: \( a = 1 \), \( b = -3 \), \( c = 2 \). 2. Calculate the discriminant: \[ \Delta = (-3)^2 - 4 \cdot 1 \cdot 2 = 9 - 8 = 1 \] 3. Apply the quadratic formula: \[ x = \frac{-(-3) \pm \sqrt{1}}{2 \cdot 1} = \frac{3 \pm 1}{2} \] 4. Simplify the solutions: \[ x_1 = \frac{3 + 1}{2} = 2 \] \[ x_2 = \frac{3 - 1}{2} = 1 \]The solutions are \( x = 2 \) and \( x = 1 \).Example 2: Zero Discriminant Solve the quadratic equation \( x^2 - 4x + 4 = 0 \).1. Identify the coefficients: \( a = 1 \), \( b = -4 \), \( c = 4 \). 2. Calculate the discriminant: \[ \Delta = (-4)^2 - 4 \cdot 1 \cdot 4 = 16 - 16 = 0 \] 3. Apply the quadratic formula: \[ x = \frac{-(-4) \pm \sqrt{0}}{2 \cdot 1} = \frac{4 \pm 0}{2} \] 4. Simplify the solution: \[ x = \frac{4}{2} = 2 \]The solution is \( x = 2 \) (a repeated root).Example 3: Negative Discriminant Solve the quadratic equation \( x^2 + 4x + 5 = 0 \).1. Identify the coefficients: \( a = 1 \), \( b = 4 \), \( c = 5 \). 2. Calculate the discriminant: \[ \Delta = 4^2 - 4 \cdot 1 \cdot 5 = 16 - 20 = -4 \] 3. Apply the quadratic formula: \[ x = \frac{-4 \pm \sqrt{-4}}{2 \cdot 1} = \frac{-4 \pm 2i}{2} \] 4. Simplify the solutions: \[ x_1 = \frac{-4 + 2i}{2} = -2 + i \] \[ x_2 = \frac{-4 - 2i}{2} = -2 - i \]The solutions are \( x = -2 + i \) and \( x = -2 - i \) (complex roots).ConclusionThe quadratic formula is a powerful tool for solving quadratic equations, providing a straightforward method to find real or complex roots. By following these steps, you can systematically solve any quadratic equation. Remember to check the discriminant first to understand the nature of the roots before applying the formula.If you have any questions or need further examples, feel free to ask!

More Examples

Example 1: Positive Discriminant

Problem: Solve the quadratic equation \( x^2 – 4x – 5 = 0 \).

Solution:

1. Identify the coefficients:
\[
a = 1, \; b = -4, \; c = -5
\]

2. Calculate the discriminant (\(\Delta\)):
\[
\Delta = b^2 – 4ac
\]
\[
\Delta = (-4)^2 – 4 \cdot 1 \cdot (-5)
\]
\[
\Delta = 16 + 20
\]
\[
\Delta = 36
\]

3. Apply the quadratic formula:
\[
x = \frac{-b \pm \sqrt{\Delta}}{2a}
\]
\[
x = \frac{-(-4) \pm \sqrt{36}}{2 \cdot 1}
\]
\[
x = \frac{4 \pm 6}{2}
\]

4. Simplify the solutions:
\[
x_1 = \frac{4 + 6}{2} = \frac{10}{2} = 5
\]
\[
x_2 = \frac{4 – 6}{2} = \frac{-2}{2} = -1
\]

Solutions: \( x = 5 \) and \( x = -1 \)

Example 2: Zero Discriminant

Problem: Solve the quadratic equation \( 2x^2 – 4x + 2 = 0 \).

Solution:

1. Identify the coefficients:
\[
a = 2, \; b = -4, \; c = 2
\]

2. Calculate the discriminant (\(\Delta\)):
\[
\Delta = b^2 – 4ac
\]
\[
\Delta = (-4)^2 – 4 \cdot 2 \cdot 2
\]
\[
\Delta = 16 – 16
\]
\[
\Delta = 0
\]

3. Apply the quadratic formula:
\[
x = \frac{-b \pm \sqrt{\Delta}}{2a}
\]
\[
x = \frac{-(-4) \pm \sqrt{0}}{2 \cdot 2}
\]
\[
x = \frac{4 \pm 0}{4}
\]
\[
x = \frac{4}{4} = 1
\]

Solution: \( x = 1 \) (a repeated root)

Example 3: Negative Discriminant

Problem: Solve the quadratic equation \( x^2 + 2x + 5 = 0 \).

Solution:

1. Identify the coefficients:
\[
a = 1, \; b = 2, \; c = 5
\]

2. Calculate the discriminant (\(\Delta\)):
\[
\Delta = b^2 – 4ac
\]
\[
\Delta = 2^2 – 4 \cdot 1 \cdot 5
\]
\[
\Delta = 4 – 20
\]
\[
\Delta = -16
\]

3. Apply the quadratic formula:
\[
x = \frac{-b \pm \sqrt{\Delta}}{2a}
\]
\[
x = \frac{-2 \pm \sqrt{-16}}{2 \cdot 1}
\]
\[
x = \frac{-2 \pm 4i}{2}
\]

4. Simplify the solutions:
\[
x_1 = \frac{-2 + 4i}{2} = -1 + 2i
\]
\[
x_2 = \frac{-2 – 4i}{2} = -1 – 2i
\]

Solutions: \( x = -1 + 2i \) and \( x = -1 – 2i \) (complex roots)