How to solve quadratic equations

The history of quadratic equations goes as far back as Babylonian times, and the fact that it is still taught to this day proves that it has real-world applications.

These days, students will typically come across quadratic equations in their high school algebra class. It forms a large part of the national curriculum, which is why understanding it is so important.

But what exactly is it, and how do you solve them? That’s what we’ll explain in this article.

In algebra, a quadratic equation is an equation that can be written in standard form where x is an unknown variable, and a, b, and c are known values. Also, a cannot be 0, and c is a constant term. The standard form of a quadratic equation looks something like this:

ax² + bx + c = 0

It is called a quadratic equation because the word ‘quad’ means squared, in reference to the unknown variable x, which gets squared. It should be noted that the values of x that solve a quadratic equation are known as its solutions.

Examples of quadratic equations

Generally speaking, quadratic equations follow the following formula: ax² + bx + c = 0.

But in reality, not all quadratic equations will look like that. There are different variations of this equation that you’ll want to be aware of. Let’s take a look at a few examples.

Example 1: 3x² + 9x + 2 = 0

3x² + 9x + 2 = 0 is a quadratic equation because in this example:

• a = 3
• b = 9
• c = 2.

This is an easy one since it follows the exact quadratic equation formula. What about the next one?

Example 2: x² - 6x = 0

This may be a little difficult to figure out, but x² - 6x = 0 is also a quadratic equation. In this example:

• a = 1 but we don’t write 1x², instead, we simply leave it as x²
• b = - 6
• c = 0 and is therefore removed from the equation entirely.

Example 3: x² = 5x - 4

At first glance, this doesn’t look like a quadratic equation, but it’s a quadratic equation in disguise.

If you move all terms to the left of the equal sign, the equation will be rearranged to x² - 5x + 4. In this example:

• a = 1 but we don’t write 1x², instead, we simply leave it as x2
• b = - 5
• c = 4

Example 4: 3x − 4 = 0

3x − 4 = 0 is not a quadratic equation. Why? Because it’s missing the ‘quad’ part of the equation, i.e., the x².

The main method to determine if a quadratic has a solution is to look at its discriminant:

• If the discriminant is positive, the quadratic equation has two solutions.
• If the discriminant is equal, the quadratic equation has one solution.
• If the discriminant is negative, the quadratic equation has no solution.

The discriminant is represented by the equation b² – 4ac. This is derived from using the quadratic formula, which is:

[-b ± √(b² - 4ac)] / (2a)

As you can see, the discriminant is the expression beneath the radical sign (square root). Let’s see the discriminant in action.

Example 1: x² + 12x + 36

In this example, a = 1, b = 12, and c = 36. We can substitute these values into the discriminant to see if it has any solutions.

• The discriminant is as follows: b² – 4ac.
• Substituting in our a, b, and c values gives us the equation: 122 - 4(1)(36).
• Calculating this equation gives us: 144 - 144 = 0.

Therefore, b² – 4ac = 0. We know that if the discriminant is equal, the quadratic equation has one real solution.

This means that x² + 12x + 36 has one real solution.

Example 2: x² + 2x + 15

In this example, a = 1, b = 2, and c = 15. We can substitute these values into the discriminant to see if it has any solutions.

• The discriminant is as follows: b² – 4ac.
• Substituting in our a, b, and c values gives us the equation: 22 - 4(1)(15).
• Calculating this equation gives us: 4 - 60 = - 54.

Therefore, b² – 4ac < 0. We know that if the discriminant is a negative number, the quadratic equation has no real solution.

This means that x² + 2x + 15 has no real solution.

There are three methods you can use when solving quadratic equations – factorization, the quadratic formula, and completing the square. It should be noted that we are only solving for real solutions, not complex solutions.

Factorization method

The factorization method is as follows:

• Step 1: Combine all terms to one side of the equation.
• Step 2: Multiply a and c.
• Step 3: Find the two numbers that multiply to create the value (a x c), and add or subtract to create b.
• Step 4: Rearrange the quadratic equation with the factors.
• Step 5: Calculate x to reach the final answer.
• Step 6: Substitute the x values to check our answer.

Sounds easy enough, right? Let’s see this as an example.

Example 1: x² + 10x + 21 = 0

To solve x² + 10x + 21 = 0, let’s follow the steps outlined above.

• Step 1: All terms are already on one side of the equation.
• Step 2: Multiplying a (1) and c (21) gives us 21.
• Step 3: Now, we need to find two numbers that multiply to create 21 and add or subtract to create 10. Using our knowledge of times tables, we know that 7 and 3 multiply to make 21 and add to make 10.
• Step 4: Rearranging the quadratic equation with the factors gives us (x + 7)(x + 3) = 0.
• Step 5: Calculating for x means that x = -7 and x = -3.
• Step 6a: Substituting -7 into the quadratic equation gives us: (-7)² + 10(-7) + 21 = 49 - 70 + 21 = 0.
• Step 6b: Substituting -3 into the quadratic equation gives us: (-3)² + 10(-3) + 21 = 9 - 30 + 21 = 0.

The quadratic equation equals zero with both factors, which means that both (x = -7) and (x = -3) are a real solution to the problem.

Example 2: x² - 4x = 45

To solve x² - 4x = 45, let’s follow the steps outlined above.

• Step 1: Combining all terms to one side of the equation gives us: x² - 4x - 45 = 0.
• Step 2: Multiplying a (1) and c (-45) gives us -45.
• Step 3: Now we need to find two numbers that multiply to create -45 and add or subtract to create -4. Using our knowledge of times tables, we know that -9 and 5 multiply to make -45 and add to make -4.
• Step 4: Rearranging the quadratic equation with the factors gives us (x - 9)(x + 5) = 0
• Step 5: Calculating for x means that x = 9 and x = -5.
• Step 6a: Substituting 9 into the quadratic equation gives us: 9² - 4(9) - 45 = 81 - 36 - 45 = 0.
• Step 6b: Substituting -5 into the quadratic equation gives us: -5² - 4(-5) - 45 = 25 - (-20) - 45 = 0.

The quadratic equation equals zero with both factors, which means that both (x = 9) and x = -5) are a real solution to the problem.

Quadratic formula

Here is how to solve quadratic equations using the quadratic formula:

• Step 1: Combine all terms to one side of the equation.
• Step 2: Substitute a, b, and c into the quadratic formula: [-b ± √(b² - 4ac)] / (2a) and solve to find the final answer.
• Step 3: Substitute the x values to check the answer.

Example 1: x² + 6x - 7 = 0

To solve x² + 6x + 7 = 0, let’s follow the steps outlined above.

• Step 1: All terms are already combined to one side so we can move on to step 2
• Step 2a: a = 1, b = 6, c = -7. Substituting these into the quadratic formula gives us [-6 ± √(6² - 4(1)(-7)] / (2(1).
• Step 2b: This simplifies to [-6 ± √(64)] / 2.
• Step 2c: This again simplifies to [-6 ± 8] / 2. Calculating this formula means that (x = -7) or (x = 1).
• Step 3a: Substituting -7 into the quadratic equation gives us: -7² + 6(-7) - 7 = 49 - 42 - 7 = 0.
• Substituting 1 into the quadratic equation gives us: 1² + 6(1) - 7 = 1 + 6 - 7 = 0.

The quadratic equation equals zero with both factors, which means that both (x = -7) and x = 1) are a real solution to the problem.

Completing the square

The last method is called completing the square. Here’s how it’s done:

• Step 1: Rearrange the equation in the form ax² + bx = – c
• Step 2: Divide both sides by a (if a = 1, skip this step).
• Step 3: Divide b by two and square it (b/2)², do this to both sides.
• Step 4: Find the perfect square and simplify.
• Step 5: Find the square root of both sides.
• Step 6: Simplify and solve.

It can be a little tricky to visualize, so let’s use the same quadratic equation as the previous example since we already know the answer to it.

Example 1: x² + 6x - 7 = 0

To solve x2 + 6x - 7, let’s follow the steps outlined above.

• Step 1: Rearrange the equation in the form ax² + bx = – c. This gives us x² + 6x = 7.
• Step 2: a = 1, which means we can skip this step.
• Step 3: First, we need to divide b by two and square it. (b/2)² = (6/2)² = 32 = 9. We can add this to both sides of the equation to give us x² + 6x + 9 = 7 + 9.
• Step 4: Now we can find the perfect square of x² + 6x + 9 = (x + 2)². This means we can simplify the entire equation to (x + 3)² = 16.
• Step 5: Square rooting both sides of the equation gives us √(x + 3)² = √16.
• Step 6: Simplifying this equation gives us x + 3 = ± 4. Solving it means that x = -3 ± 4. As a result, (x = 1) or (x = -7). These are our final two solutions.

That was a slightly easy example, but what if the a coefficient does not equal 1?

Example 2: 5x² - 10x - 15 = 0

To solve 5x² - 10x - 15, let’s follow the steps outlined above.

• Step 1: Rearrange the equation in the form ax² + bx = – c. This gives us 5x² - 10x = 15
• Step 2: a = 5, which means we must divide the equation by 5 to give us x² - 2x = 3
• Step 3: First, we need to divide b by two and square it. (b/2)² = (-2/2)² = -12 = 1. We then add this to both sides of the equation to give us x² - 2x +1 = 3 + 1
• Step 4: Now we can find the perfect square of x² - 2x + 1 = (x - 1)². This means we can simplify the entire equation to (x - 1)² = 4
• Step 5: Square rooting both sides of the equation gives us √(x - 1)² = √4.
• Step 6: Simplifying this equation gives us x - 1 = ± 2. Solving it means that x = 1 ± 2. As a result, (x = 3) or (x = -1). These are our final two solutions.

The quadratic equation is an equation that has an x² and is in the standard form ax² + bx + c = 0. You can determine whether a quadratic equation has a real solution or not by using the discriminant b² – 4ac.

However, to find the solutions, you will have to use one of three methods – factorization, the quadratic formula, or completing the square.