Quadratic Equations

Solve $x^2 = 5x$

Wrong: Divide both sides by $x$: $x = 5$

Students divide both sides by $x$, not realising that $x = 0$ is also a valid solution. Dividing by a variable can "lose" solutions.

Never divide by the variable — it might be zero!

Instead, rearrange and factorise: $$x^2 = 5x$$ $$x^2 - 5x = 0$$ $$x(x - 5) = 0$$ $$x = 0 \text{ or } x = 5$$

You had two solutions, but dividing by $x$ lost the $x = 0$ case.

Solve $x^2 - 9 = 0$

Wrong: $x^2 = 9$ $x = 3$

Students find one square root and stop. They forget that both positive and negative values square to give a positive result.

When you take a square root, always consider both the positive and negative roots.

$x^2 = 9$ $x = \pm\sqrt{9}$ $x = 3$ or $x = -3$

Check: $3^2 = 9$ ✓ and $(-3)^2 = 9$ ✓

Memory aid: "Square roots come in pairs — positive and negative!"

Solve $x^2 - 4x - 5 = 0$ using the quadratic formula.

Wrong: $a = 1, b = 4, c = -5$ $x = \frac{-4 \pm \sqrt{16 + 20}}{2}$

Students misread the coefficient $b$ when the term is negative. In $x^2 - 4x - 5$, we have $b = -4$, not $b = 4$.

Write out $a$, $b$, $c$ carefully, including signs.

For $x^2 - 4x - 5 = 0$:

• $a = 1$
• $b = -4$ (the coefficient of $x$, including the negative)
• $c = -5$

$$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-5)}}{2(1)}$$ $$x = \frac{4 \pm \sqrt{16 + 20}}{2} = \frac{4 \pm 6}{2}$$ $$x = 5 \text{ or } x = -1$$

Solve $x^2 + 3x = 10$

Wrong: $x(x + 3) = 10$ So $x = 10$ or $x + 3 = 10$, giving $x = 7$

Students try to factorise before rearranging to equal zero. They wrongly apply the "if $ab = 0$ then $a = 0$ or $b = 0$" rule to non-zero products.

The zero product rule only works when the product equals zero.

Correct method: $$x^2 + 3x = 10$$ $$x^2 + 3x - 10 = 0$$ $$(x + 5)(x - 2) = 0$$ $$x = -5 \text{ or } x = 2$$

Check: $(-5)^2 + 3(-5) = 25 - 15 = 10$ ✓