Solving Quadratic Equations

First rearrange so one side is zero: $ax^2 + bx + c = 0$. This is essential, because the null factor law only applies to a product equal to zero.

Factorise the left side, then set each factor to zero. From $x^2 - 5x + 6 = 0$ you get $(x-2)(x-3) = 0$, so $x = 2$ or $x = 3$.

When there is no constant term, factorise out the variable: $x^2 = 5x$ becomes $x^2 - 5x = 0$, so $x(x-5) = 0$, giving $x = 0$ or $x = 5$.

Rearrange to make one side zero, factorise, then apply the null factor law: each factor set to zero gives a solution. Never divide both sides by the variable, because that loses the solution $x = 0$.

Pause, copy the null factor law, the worked $x^2 - 5x + 6 = 0$ giving $x = 2$ or $x = 3$, and the warning about dividing by $x$, into your book.

We just saw that factorisation is fast when the numbers are friendly. That raises a question: what do you do when the quadratic does not factorise over the integers? This card answers it → completing the square rewrites any quadratic so the variable appears only once, and then you can just take a square root.

Halve the coefficient of $x$ and square it. For $x^2 + 6x$, half of 6 is 3, and $3^2 = 9$, so $x^2 + 6x = (x+3)^2 - 9$. You subtract the 9 you added.

For $x^2 + 6x - 3 = 0$: rewrite as $(x+3)^2 - 9 - 3 = 0$, so $(x+3)^2 = 12$.

Now take the square root of both sides, remembering both signs: $x + 3 = \pm\sqrt{12} = \pm 2\sqrt{3}$, so $x = -3 \pm 2\sqrt{3}$.

Complete the square by halving the coefficient of $x$, squaring it, adding and subtracting it. Then take the square root of both sides and keep the $\pm$. Divide through by $a$ first if $a \neq 1$.

Pause, copy the halve-and-square method, the worked $x^2 + 6x - 3 = 0$ giving $x = -3 \pm 2\sqrt{3}$, and the reminder to keep both signs when square rooting, into your book.

We just saw that completing the square works on any quadratic but takes several steps every time. That raises a question: can that work be done once, in general, so you never repeat it? This card answers it → that is exactly what the quadratic formula is.

Completing the square on the general equation $ax^2 + bx + c = 0$ produces $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. The formula is that one calculation, done once and reused.

Substitute carefully, keeping the sign of each coefficient. For $2x^2 - 4x - 1 = 0$: $a = 2$, $b = -4$, $c = -1$, so $x = \dfrac{4 \pm \sqrt{16 + 8}}{4} = \dfrac{4 \pm 2\sqrt{6}}{4} = \dfrac{2 \pm \sqrt{6}}{2}$.

Choosing a method: try factorising for a few seconds. If it does not come, use the formula. Use completing the square when the question asks for it, or when you want the vertex form.

The quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ is completing the square done once in general, so it solves any quadratic. Substitute with the signs attached, simplify the surd, and cancel common factors.

Pause, copy the quadratic formula, the worked $2x^2 - 4x - 1 = 0$ giving $\frac{2 \pm \sqrt{6}}{2}$, and the method-choice rule (factorise first, formula if it does not come), into your book.

1. Solve $x^2 - 9 = 0$.
2. Solve $x^2 + 7x + 12 = 0$.
3. Solve $x^2 = 4x$.
4. Solve $x^2 + 2x - 5 = 0$ in exact form.