Solving Quadratic Equations

Welcome, students 👋 Today you will learn how to solve quadratic equations and why they matter in functions. Quadratics appear everywhere: the path of a basketball, the shape of a bridge arch, and even profit models in business. By the end of this lesson, you should be able to explain what a quadratic equation is, solve it using several methods, and connect the solutions to graphs and function behavior.

Learning objectives:

Explain the key ideas and terminology behind solving quadratic equations.
Use standard IB methods to solve quadratic equations accurately.
Connect quadratic equations to the study of functions and graphs.
Recognize when a quadratic has two, one, or no real solutions.
Use examples to justify answers clearly and logically.

What is a quadratic equation?

A quadratic equation is an equation that can be written in the form $ax^2+bx+c=0$, where $a\ne 0$. The expression $ax^2+bx+c$ is called a quadratic polynomial. The highest power of the variable is $2$, which is why it is called “quadratic.”

When you solve a quadratic equation, you are finding the value or values of $x$ that make the equation true. These values are called the roots, solutions, or zeros of the equation. In function language, if $f(x)=ax^2+bx+c$, then the solutions of $f(x)=0$ are the $x$-intercepts of the graph of the function.

For example, the equation $x^2-5x+6=0$ is quadratic. The solutions are $x=2$ and $x=3$ because both values make the expression equal to $0$. On a graph, the parabola $y=x^2-5x+6$ crosses the $x$-axis at $(2,0)$ and $(3,0)$. 📈

This connection between equations and graphs is very important in IB Mathematics: Analysis and Approaches HL. Solving a quadratic is not only about algebra. It also tells you where a function touches or crosses the horizontal axis.

Method 1: Factorising the quadratic

One of the quickest ways to solve a quadratic equation is by factorising. The idea is to rewrite the quadratic as a product of two brackets and then use the zero-product property. This property says: if $ab=0$, then $a=0$ or $b=0$.

Let’s solve $x^2-5x+6=0$.

We factorise:

$$x^2-5x+6=(x-2)(x-3)$$

So the equation becomes:

$$(x-2)(x-3)=0$$

Using the zero-product property:

$$x-2=0 \quad \text{or} \quad x-3=0$$

Therefore:

$$x=2 \quad \text{or} \quad x=3$$

This method works best when the quadratic factors neatly with integers or simple fractions. In exams, it is always worth checking whether factorising is possible first because it is often the fastest method.

Another example is $2x^2+7x+3=0$.

We factorise:

$$2x^2+7x+3=(2x+1)(x+3)$$

Then:

$$(2x+1)(x+3)=0$$

So:

$$2x+1=0 \quad \text{or} \quad x+3=0$$

Giving:

$$x=-\frac{1}{2} \quad \text{or} \quad x=-3$$

Always check your answers by substituting them back into the original equation. This is a strong IB habit because it reduces careless errors. ✅

Method 2: Completing the square

Not every quadratic factorises easily. In those cases, completing the square is a powerful method. It also helps you understand the vertex form of a parabola.

The goal is to rewrite $ax^2+bx+c$ into a form like $a(x-h)^2+k$. This reveals the turning point of the graph and can help solve the equation.

Let’s solve $x^2+6x+5=0$ by completing the square.

First, move the constant term:

$$x^2+6x=-5$$

Take half of the coefficient of $x$, which is $6$, so half is $3$. Then square it to get $9$.

Add $9$ to both sides:

$$x^2+6x+9=4$$

Rewrite the left side as a perfect square:

$$(x+3)^2=4$$

Now take the square root of both sides:

$$x+3=\pm 2$$

So:

$$x=-3\pm 2$$

Therefore the two solutions are:

$$x=-1 \quad \text{or} \quad x=-5$$

Completing the square is especially useful in more advanced work because it links algebra to graph transformations. For example, the function $f(x)=x^2+6x+5$ can be rewritten as $f(x)=(x+3)^2-4$, which shows that the parabola has vertex $(-3,-4)$. That information helps you sketch the graph and understand its minimum value.

Method 3: The quadratic formula

The quadratic formula is a general method that works for every quadratic equation. If $ax^2+bx+c=0$, then the solutions are given by:

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

This formula is extremely important because it solves any quadratic, even when factorising is difficult or impossible.

Let’s solve $3x^2-4x-2=0$.

Here, $a=3$, $b=-4$, and $c=-2$.

Substitute into the formula:

$$x=\frac{-(-4)\pm\sqrt{(-4)^2-4(3)(-2)}}{2(3)}$$

Simplify:

$$x=\frac{4\pm\sqrt{16+24}}{6}$$

$$x=\frac{4\pm\sqrt{40}}{6}$$

Since $\sqrt{40}=2\sqrt{10}$, we get:

$$x=\frac{4\pm2\sqrt{10}}{6}$$

$$x=\frac{2\pm\sqrt{10}}{3}$$

So the exact solutions are:

$$x=\frac{2+\sqrt{10}}{3} \quad \text{and} \quad x=\frac{2-\sqrt{10}}{3}$$

A key feature of the quadratic formula is the expression under the square root, $b^2-4ac$. This is called the discriminant. It tells you how many real solutions the quadratic has.

If $b^2-4ac>0$, there are two real solutions.
If $b^2-4ac=0$, there is one real solution, or a repeated root.
If $b^2-4ac<0$, there are no real solutions.

For example, in $x^2+4x+5=0$, we have $a=1$, $b=4$, and $c=5$, so

$$b^2-4ac=16-20=-4$$

Because this is negative, the equation has no real solutions. On the graph, the parabola does not cross the $x$-axis.

Quadratic equations in function language

In functions, solving $f(x)=0$ means finding where the graph of $y=f(x)$ meets the horizontal axis. This is a central idea in the topic of Functions.

Suppose $f(x)=x^2-2x-3$. To find the zeros, solve:

$$x^2-2x-3=0$$

Factorising gives:

$$(x-3)(x+1)=0$$

So the zeros are:

$$x=3 \quad \text{and} \quad x=-1$$

This means the graph crosses the $x$-axis at $(3,0)$ and $(-1,0)$. If the function represents height, these roots could mean when a ball is on the ground. If the function represents profit, the roots could mean the break-even points. 💡

Quadratic equations also appear when comparing two functions. If $f(x)=g(x)$ and both are quadratic or one leads to a quadratic after rearranging, then the solutions are the $x$-values where the graphs intersect. For example, if $f(x)=x^2$ and $g(x)=2x+3$, solving $f(x)=g(x)$ gives

$$x^2=2x+3$$

which becomes

$$x^2-2x-3=0$$

The solutions are $x=-1$ and $x=3$, so the graphs intersect at two points.

This is a powerful interpretation because IB often asks you to move between algebraic and graphical representations. A solved equation should always make sense on the graph too.

Common mistakes and how to avoid them

A very common mistake is forgetting to put the equation into the form $ax^2+bx+c=0$ before solving. Always move all terms to one side first.

Another mistake is sign errors, especially when using the quadratic formula or completing the square. For example, if $b=-4$, then $-b=4$, not $-4$. Careful substitution matters.

Students also sometimes forget that a quadratic can have two equal solutions. For example,

$$x^2-6x+9=0$$

factorises as

$$(x-3)^2=0$$

so the only solution is

$$x=3$$

This repeated root means the graph just touches the $x$-axis at one point instead of crossing it.

Finally, always check whether your answers are exact or approximate. In IB, exact answers are often preferred when possible, such as $\frac{2\pm\sqrt{10}}{3}$ instead of decimal values.

Conclusion

Solving quadratic equations is a core skill in the study of functions because it connects algebra, graphs, and real-life modeling. You can solve quadratics by factorising, completing the square, or using the quadratic formula. Each method has its place, and each answer tells you something about the graph of the function. The roots of a quadratic are the $x$-intercepts, and the discriminant tells you how many real solutions exist.

For IB Mathematics: Analysis and Approaches HL, the goal is not just to get an answer, but to explain the method clearly, interpret the result, and connect it to the function representation. When students sees a quadratic equation, the key question is: which method is most efficient, and what does the solution mean in context? 🎯

Study Notes

A quadratic equation has the form $ax^2+bx+c=0$, where $a\ne 0$.
The solutions are also called roots, zeros, or x-intercepts.
Factorising uses the zero-product property: if $ab=0$, then $a=0$ or $b=0$.
Completing the square rewrites a quadratic into the form $a(x-h)^2+k$.
The quadratic formula is $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.
The discriminant is $b^2-4ac$.
If $b^2-4ac>0$, there are two real solutions.
If $b^2-4ac=0$, there is one repeated real solution.
If $b^2-4ac<0$, there are no real solutions.
In function language, solving $f(x)=0$ means finding the $x$-intercepts of the graph of $y=f(x)$.
Solving $f(x)=g(x)$ finds points where two graphs intersect.
Always check your solutions by substitution or graph interpretation.