Solving Quadratic Equations

Hello! Welcome to the fourth lesson in your Mathematics Mastery Sprint.

In our last lesson, we built a solid strategy for solving linear equations, even when they involved tricky algebraic fractions. We focused on the systematic process of expanding brackets, clearing denominators, and isolating the variable.

Today, we take a significant step up to tackle quadratic equations. These are equations of the form $a x^{2} + b x + c = 0$ , and they are a cornerstone of the IGCSE paper. Mastering them is essential for hitting your target grade.

Our learning outcome for this lesson is to solve quadratic equations by factorization, using the quadratic formula, and by completing the square. You've mentioned that you sometimes choose inefficient methods, so our goal today is not just to learn how to use these three techniques, but to understand the mechanics of each one perfectly. This will set the stage for our next lesson, where we'll focus on strategically choosing the best method for any given problem.

1. Solving by Factorization

This is often the fastest method, but it only works if the quadratic expression can be factorized into whole numbers or simple fractions. The method relies on a simple but powerful idea called the Zero Product Property: if the product of two numbers is zero, then at least one of those numbers must be zero. If $A \times B = 0$ , then either $A = 0$ or $B = 0$ .

The process is as follows:

Rearrange the equation into the standard form $a x^{2} + b x + c = 0$ .
Factorize the quadratic expression on the left-hand side.
Set each factor (i.e., each bracket) equal to zero.
Solve the two resulting linear equations to find the two possible values for $x$ .

The following resource provides a clear, step-by-step guide with excellent worked examples.

Quadratic Equations | CIE IGCSE Maths

This guide from Save My Exams is tailored for your IGCSE syllabus. It clearly explains the factorisation method and includes a crucial warning about a common mistake.

Please read the first section, 'Solving Quadratics by Factorising' (pages 1-5). Pay close attention to the examples where the coefficients are not 1, and make sure you understand the warning: 'It is a common mistake to divide both sides by x at the beginning - you will lose a solution'.

This image provides a great visual summary of the process for three different types of quadratics.

How to Factor Quadratic Equations — This image shows three worked examples of solving quadratic equations by factorization. Each example shows the equation being factorized, and then each factor being set to zero to find the solutions for x.

Test your understanding!

Solve the equation $2 x^{2} - 5 x = 3$ .

Show answer

Rearrange to the form $a x^{2} + b x + c = 0$ :
$2 x^{2} - 5 x - 3 = 0$
Factorize the quadratic. We are looking for two numbers that multiply to give $2 \times - 3 = - 6$ and add to give $- 5$ . These numbers are $- 6$ and $+ 1$ . We use these to split the middle term:
$2 x^{2} - 6 x + 1 x - 3 = 0$
Now, factor by grouping:
$2 x (x - 3) + 1 (x - 3) = 0$
$(2 x + 1) (x - 3) = 0$
Set each factor to zero and solve: $2 x + 1 2 x x = = = 0 - 1 - \frac{1}{2} or x - 3 x = = 03$ The two solutions are $x = - \frac{1}{2}$ and $x = 3$ .

2. Solving with the Quadratic Formula

What happens when a quadratic doesn't factorize nicely? The quadratic formula is your universal tool. It can solve any quadratic equation. It might seem intimidating, but it's just a matter of careful substitution.

For any equation in the form $a x^{2} + b x + c = 0$ , the solutions are given by:
$x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$

Key steps:

Ensure the equation is in the form $a x^{2} + b x + c = 0$ .
Carefully identify the values of $a$ , $b$ , and $c$ . Pay close attention to negative signs!
Substitute these values into the formula. It's good practice to use brackets for negative numbers, e.g., if $b = - 4$ , write $(- 4)^{2}$ .
Calculate the two solutions, one using the '+' and one using the '-'.

This video demonstrates the process clearly and shows how to simplify the result.

Solving Quadratic Equations (5 Methods)

This segment from Mario's Math Tutoring gives a concise demonstration of applying the quadratic formula.

Watch the section on the Quadratic Formula (02:06 - 04:28). Notice how he identifies a, b, and c, substitutes them into the formula, and simplifies the resulting surd. This is a key skill for non-calculator papers or when 'exact answers' are required.

The Discriminant: A Clue to the Solution

The part of the formula under the square root, $b^{2} - 4 a c$ , is called the discriminant. Its value gives you a vital clue about the solutions without having to do the full calculation:

If $b^{2} - 4 a c > 0$ , there are two distinct real solutions. (The graph crosses the x-axis twice).
If $b^{2} - 4 a c = 0$ , there is one repeated real solution. (The vertex of the graph touches the x-axis).
If $b^{2} - 4 a c < 0$ , there are no real solutions. (The graph does not cross the x-axis).

This is a powerful concept we will explore further in the next lesson to help choose the best solving method.

Test your understanding!

Solve the equation $3 x^{2} + 5 x - 1 = 0$ , giving your answers to 2 decimal places.

Show answer

Identify a, b, c: Here, $a = 3$ , $b = 5$ , and $c = - 1$ .
Substitute into the formula:
$x = \frac{- ( 5 ) \pm ( 5 ) ^{2} - 4 ( 3 ) ( - 1 )}{2 ( 3 )}$
Simplify:
$x = \frac{- 5 \pm 25 - ( - 12 )}{6}$
$x = \frac{- 5 \pm 37}{6}$
Calculate the two solutions:
$x = \frac{- 5 + 37}{6} \approx \frac{- 5 + 6.082...}{6} \approx 0.180... \approx 0.18$
$x = \frac{- 5 - 37}{6} \approx \frac{- 5 - 6.082...}{6} \approx - 1.847... \approx - 1.85$
The solutions are $x = 0.18$ and $x = - 1.85$ .

3. Solving by Completing the Square

This method is less common for direct solving unless a question specifically asks for it. However, the skill of "completing the square" is vital for finding the turning point of a quadratic graph (a topic you find challenging) and for rearranging complex formulas.

The goal is to transform one side of the equation into a perfect square trinomial, i.e., $(x + p)^{2}$ .

Key steps for solving $x^{2} + b x + c = 0$ :

Move the constant term ( $c$ ) to the right side: $x^{2} + b x = - c$ .
Take half of the coefficient of $x$ (which is $b$ ), square it $(\frac{b}{2})^{2}$ , and add it to both sides.
The left side now factors into a perfect square: $(x + \frac{b}{2})^{2}$ .
Solve for $x$ by taking the square root of both sides (remembering the $\pm$ symbol) and then isolating $x$ .

If the coefficient $a$ is not 1, you must divide the entire equation by $a$ before you begin.

The Save My Exams guide provides an excellent text-based explanation.

Quadratic Equations | CIE IGCSE Maths

This section of the Save My Exams guide breaks down both how to complete the square and how to use it to solve an equation.

Read pages 13-15 on 'Solving by Completing the Square'. Focus on the worked example for 2x^2 - 8x - 24 = 0, as it shows the crucial first step of dividing by 'a'.

Here is a visual step-by-step example of the process.

Solving Quadratic Equations by Completing the Square — This image shows a clear, step-by-step solution for the equation x² - 10x - 12 = 0 using the completing the square method. It demonstrates moving the constant, adding the squared half-coefficient to both sides, factoring the perfect square, and solving for x to get an exact answer in surd form.

Test your understanding!

A question asks:
(a) Write $x^{2} - 6 x + 2$ in the form $(x + p)^{2} + q$ .
(b) Hence, solve the equation $x^{2} - 6 x + 2 = 0$ , giving your answer in exact surd form.

Show answer

(a) Completing the square:

Focus on $x^{2} - 6 x$ . Half of -6 is -3.
Rewrite as $(x - 3)^{2} - (- 3)^{2}$ . This is $(x - 3)^{2} - 9$ .
Now include the original +2: $(x - 3)^{2} - 9 + 2$ .
The final form is $(x - 3)^{2} - 7$ .

(b) Solving the equation:

The word "Hence" means we must use the result from part (a).
Set the completed square form to zero:
$(x - 3)^{2} - 7 = 0$
Rearrange to solve for $x$ :
$(x - 3)^{2} = 7$
$x - 3 = \pm 7$
$x = 3 \pm 7$
The two exact solutions are $x = 3 + 7$ and $x = 3 - 7$ .

Conclusion

Today we've armed you with three distinct methods for solving any quadratic equation you might encounter.

Key Takeaways:

Factorization: Use this when you can quickly spot the factors. It's the fastest method but doesn't always work.
Quadratic Formula: Your reliable, all-purpose tool. It works for every quadratic equation and is essential when answers are required to a certain accuracy or in surd form.
Completing the Square: A fundamental algebraic skill that is often tested explicitly and is crucial for understanding the properties of quadratic graphs.

You now have the complete toolkit for solving quadratics. In our next lesson, we will focus entirely on strategy. We will look at exam-style questions and practice deciding which of these three methods is the most efficient and least error-prone choice for each one. This will be a critical step in boosting your speed and confidence.

Recommended resources

1. Solving by Factorization

2. Solving with the Quadratic Formula

The Discriminant: A Clue to the Solution

3. Solving by Completing the Square

Conclusion