- After covering quadratic expressions and functions, you can easily build connections to quadratic equations.
- Solving a quadratic means finding the value(s) of the variable that make the equation true.
- These solutions are also called the roots or zeros of the quadratic.
Quadratic equation
An equation that can be written in the form $ax^2+bx+c=0$ where $a\neq 0$.
Put The Equation Into Standard Form First
- Before using any method, it is essential to rearrange the equation so that one side equals 0.
- This makes the structure $ax^2+bx+c=0$ visible and lets you identify $a$, $b$, and $c$.
- $x^2=x+5 \;\Rightarrow\; x^2-x-5=0$
- $2x-x^2=1 \;\Rightarrow\; -x^2+2x-1=0$ (or multiply by $-1$ to get $x^2-2x+1=0$)
- $2m^2+23=14m \;\Rightarrow\; 2m^2-14m+23=0$
- A very common mistake is to apply a method (especially the quadratic formula) before rearranging to "$=0$".
- If you do that, you will often choose incorrect values for $a$, $b$, and $c$.
Choose A Method: Factorising, Completing The Square, Or The Quadratic Formula
There are three equivalent methods commonly used in IB MYP/standard mathematics:
- Factorization (fastest when it works)
- Completing the square (powerful, often gives exact results)
- Quadratic formula (works for any quadratic, often most direct)
Which is "best" depends on the equation and what the question asks (exact answers, decimals, etc.).
- If the question asks for answers "in radical form" or "exact", avoid rounding.
- If it asks for answers to a given accuracy (e.g., 3 s.f. or 1 d.p.), the quadratic formula is often the most efficient.
Factorization Finds Solutions By Using The Zero Product Property
If you can rewrite a quadratic as a product, you can solve quickly.
Solve $x^2-7x=3$.
Solution
- Rearrange: $x^2-7x-3=0$
- Try to factorize.
- Here it does not factor nicely into integers, so factorisation is not convenient.
- But if you had, for instance: $$x^2-2x+1=0$$ $$\implies x^2-2x+1=(x-1)^2$$
- So $(x-1)^2=0 \Rightarrow x=1$ (a repeated solution).
Factorization is quickest when $a=1$ and $c$ is small, but many quadratics (especially with non-integer roots) will not factorise neatly.
Completing The Square Makes A Perfect Square You Can Solve
Completing the square rewrites a quadratic into a form like $(x+p)^2=q$, which you can solve by taking square roots.
Completing the square
A method of rewriting a quadratic into a perfect-square form (plus or minus a constant), allowing solutions by taking square roots.
Solve $x^2-4x=-1$ by completing the square.
Solution
- Add $\left(\frac{-4}{2}\right)^2=4$ to both sides: $$x^2-4x+4=-1+4$$
- Rewrite left side as a perfect square: $$(x-2)^2=3$$
- Take square roots: $$x-2=\pm\sqrt{3} \Rightarrow x=2\pm\sqrt{3}$$
To complete the square for $x^2+bx$, add $\left(\frac{b}{2}\right)^2$.
The Quadratic Formula Always Works (When $a\neq 0$)
For any quadratic in standard form $ax^2+bx+c=0$, the solutions are $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
Solve $x^2+3x-1=0$ and give answers to 1 d.p.
Solution
Here $a=1$, $b=3$, $c=-1$.
$$x=\frac{-3\pm\sqrt{3^2-4(1)(-1)}}{2} = \frac{-3\pm\sqrt{13}}{2}$$
Approximations:
- $x=\frac{-3+\sqrt{13}}{2}\approx 0.3$
- $x=\frac{-3-\sqrt{13}}{2}\approx -3.3$
When substituting into the quadratic formula, keep careful track of signs, especially when $c$ is negative or when the quadratic starts with a negative $a$.
The Discriminant Predicts The Number Of Real Solutions
Inside the quadratic formula is the expression $$b^2-4ac$$
This is called the discriminant and is denoted by the Greek letter delta: $$\Delta=b^2-4ac$$
Discriminant (Δ)
The value $\Delta=b^2-4ac$ for $ax^2+bx+c=0$, which determines the number of real solutions.
What Δ Tells You
- If $\Delta>0$, there are 2 distinct real solutions.
- If $\Delta=0$, there is 1 real solution (a repeated root).
- If $\Delta<0$, there are no real solutions (the solutions are not real numbers).
Since the solutions to $ax^2+bx+c=0$ are exactly the x-intercepts of the graph $y=ax^2+bx+c$:
- $\Delta>0$ means the parabola crosses the x-axis twice.
- $\Delta=0$ means the parabola touches the x-axis once (tangent).
- $\Delta<0$ means the parabola does not meet the x-axis.
- Think of $\Delta$ as a "preview" of the square root part of the quadratic formula.
- If $\Delta$ is negative, you would need $\sqrt{\text{negative}}$, which is why there are no real solutions.
Using The Discriminant To Decide If A Quadratic Can Be Factorised
A quadratic with integer coefficients is factorisable into real linear factors only if it has real roots, so $\Delta$ must be non-negative.
- If $\Delta<0$, it is not factorisable over the real numbers (and you will not find real roots).
- If $\Delta\ge 0$, it might be factorisable, but you still need to check whether the roots are "nice" (for example, integers or simple fractions).
- In many school settings, "factorizable" often means factorizable with integer coefficients.
- A quadratic can have $\Delta>0$ but still not factor neatly into integers (for example, if the roots involve radicals).
Solve $x^2=x+5$, and give exact answer.
Solution
- Rearrange: $x^2-x-5=0$ so $a=1$, $b=-1$, $c=-5$.
- Use the quadratic formula: $$x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(-5)}}{2(1)}}$$ $$=\frac{1\pm\sqrt{1+20}}{2}=$$ $$\frac{1\pm\sqrt{21}}{2}$$
- So the solutions are $\displaystyle x=\frac{1+\sqrt{21}}{2}$ and $\displaystyle x=\frac{1-\sqrt{21}}{2}$.
Solve $x^2+0.6x-2.6=0$ (to 3 s.f.).
Solution
- Here $a=1$, $b=0.6$, $c=-2.6$.
$$x=\frac{-0.6\pm\sqrt{0.6^2-4(1)(-2.6)}}{2}$$
$$=\frac{-0.6\pm\sqrt{0.36+10.4}}{2}$$
$$=\frac{-0.6\pm\sqrt{10.76}}{2}$$
- Since $\sqrt{10.76}\approx 3.280$,
- $x\approx \frac{-0.6+3.280}{2}=1.34$ (3 s.f.)
- $x\approx \frac{-0.6-3.280}{2}=-1.94$ (3 s.f.)
When rounding, keep extra digits during calculation, then round only at the end.
Problem Solving With A Parameter (Using Δ)
- Sometimes you are asked to choose values so the quadratic has a certain number of solutions.
- This is where the discriminant is especially useful.
Find $k$ so that $x^2+kx+16=0$ has 1 solution.
Solution
- A quadratic has 1 real (repeated) solution when $\Delta=0$.
- Here $a=1$, $b=k$, $c=16$: $$\Delta=k^2-4(1)(16)=k^2-64$$
- Set $\Delta=0$: $$k^2-64=0 \Rightarrow k^2=64 \Rightarrow k=\pm 8$$
- So $k=8$ or $k=-8$.
- What does $\Delta<0$ tell you about x-intercepts?
- If a quadratic has exactly one x-intercept, what must be true about $\Delta$?
- Why is rearranging to "$=0$" essential before identifying $a$, $b$, and $c$?
- In many problems you will decide your method based on the equation's form and the required answer.
- Try factorisation first if the numbers are small and the quadratic looks "friendly".
- Use completing the square when you want exact surd answers or when the quadratic is close to a perfect square.
- Use the quadratic formula when factorisation is not obvious, or when the question requests decimals (e.g., 1 d.p. or 3 s.f.).
- A quick method check: compute $\Delta=b^2-4ac$.
- If $\Delta$ is not a perfect square (like 13, 21, 10.76), neat factorisation is unlikely.
- If $\Delta$ is a perfect square (like 0, 1, 4, 9, 16, 25, ...), factorisation or simple roots are more likely.
