A quadratic equation can always be written or rewritten in the form:

 

$$ \large ax^2 + bx + c = 0 $$

 

It is called a quadratic equation because there is a term where \( \large x \) is raised to the second power: \( \large ax^2 \).

If \( \large a = 0 \), it is not a quadratic equation, because \( \large 0x^2 = 0 \), and then the term disappears. What remains is a linear equation.

 

Examples of quadratic equations:

 

$$ \large 40 + 2x = 2x^2 $$

$$ \large x(x - 5) - 14 = 0 $$

 

They are both quadratic equations because they can be rewritten in the form:

 

$$ \large ax^2 + bx + c = 0 $$

 

Where \( \large a \neq 0 \).

 

 

The discriminant

It can be difficult to isolate \( \large x \) in a quadratic equation as you do in a linear equation.

That is why we use the discriminant, which both helps us find the solutions and determine how many solutions there are.

The discriminant \( \large D \) is found with this formula:

 

$$ \large D = b^2 - 4ac $$

 

• If \( \large D > 0 \), there are two solutions.
• If \( \large D = 0 \), there is one solution.
• If \( \large D < 0 \), there are no real solutions.

 

Example: Find the discriminant

We want to calculate the solutions of the equation:

 

$$ \large x(x - 5) - 14 = 0 $$

 

First we expand the parentheses:

 

$$ \large x \cdot x - 5 \cdot x - 14 = 0 \quad \Leftrightarrow $$

$$ \large x^2 - 5x - 14 = 0 $$

 

Here we see that this is a quadratic equation, so we calculate the discriminant:

 

$$ \large D = b^2 - 4ac \quad \Leftrightarrow $$

$$ \large D = (-5)^2 - 4 \cdot 1 \cdot (-14) \quad \Leftrightarrow $$

$$ \large D = 25 - (-56) \quad \Leftrightarrow $$

$$ \large D = 81 $$

 

The discriminant is positive, so there are two solutions.

 

The quadratic formula

When we have the discriminant, we can find the solutions using this formula:

 

$$ \large x = \frac{-b \pm \sqrt{D}}{2a} $$

 

\( \large \pm \) is the plus/minus sign. When there are two solutions, we calculate twice: first with plus and then with minus.

 

Solution 1 (plus):

 

$$ \large x = \frac{-(-5) + \sqrt{81}}{2 \cdot 1} $$

$$ \large x = \frac{5 + 9}{2} $$

$$ \large x = 7 $$

 

Solution 2 (minus):

 

$$ \large x = \frac{-(-5) - \sqrt{81}}{2 \cdot 1} $$

$$ \large x = \frac{5 - 9}{2} $$

$$ \large x = -2 $$

 

Check

It is always a good idea to check the result by substituting the solutions into the original equation:

 

$$ \large x(x - 5) - 14 = 0 $$

 

Check solution 1:

 

$$ \large x = 7 $$

$$ \large 7(7 - 5) - 14 = 0 \Leftrightarrow $$

$$ \large 49 - 35 - 14 = 0 $$

 

That is correct.

 

Check solution 2:

 

$$ \large x = -2 $$

$$ \large -2(-2 - 5) - 14 = 0 \Leftrightarrow $$

$$ \large 4 - (-10) - 14 = 0 $$

 

That is correct.

 

Special cases

Some quadratic equations are easier to solve because one of the terms is missing.

 

When \( \large b = 0 \):

The equation takes the form:

 

$$ \large ax^2 + c = 0 $$

 

We isolate \( \large x^2 \):

 

$$ \large x^2 = -\frac{c}{a} $$

 

Here we can take the square root and find two solutions (if the right-hand side is positive):

 

$$ \large x = \pm \sqrt{-\frac{c}{a}} $$

 

When \( \large c = 0 \):

The equation takes the form:

 

$$ \large ax^2 + bx = 0 $$

 

We can factor out \( \large x \):

 

$$ \large x(ax + b) = 0 $$

 

This gives two solutions:

 

$$ \large x = 0 \quad \text{or} \quad x = -\frac{b}{a} $$

 

This method is called factorization and can often make the problem quicker to solve.

 

Summary

• A quadratic equation has the form \( \large ax^2 + bx + c = 0 \), where \( \large a \neq 0 \).
• The discriminant \( \large D = b^2 - 4ac \) determines the number of solutions.
• The quadratic formula is \( \large x = \frac{-b \pm \sqrt{D}}{2a} \).
• If \( \large D > 0 \), there are two solutions. If \( \large D = 0 \), there is one solution. If \( \large D < 0 \), there are no real solutions.
• The results can always be checked by substituting them into the original equation.