VCE Vault

Year 10 Mathematics · Number and Algebra · 20m read

Quadratic Equations — Solving and Factorising

Ever wondered why a basketball follows a curved path when you shoot it? That's a quadratic equation in action. You'll learn three powerful ways to solve these equations—factorising (breaking them into simpler parts), completing the square (a clever rearranging trick), and using the quadratic formula (your reliable backup method). Getting comfortable with all three now means you'll handle everything from physics projectiles to VCE Methods problems with real confidence.

Learning objective: Solve quadratic equations by factorising, completing the square and using the formula.

By the end of this lesson you can:

  • I factorise monic and non-monic quadratics.
  • I complete the square.
  • I use the quadratic formula.

Factorising is finding the key to a lock. A quadratic equation is a locked door. Factorising, completing the square or the quadratic formula are three different keys — any one opens it.

Three methods to solve quadratic equations

QUADRATIC EQUATIONS — general form: $ax^2 + bx + c = 0$.

A quadratic has at most TWO solutions (roots). The discriminant tells you how many.

### Method 1 — FACTORISING

Monic ($a = 1$): find $p$ and $q$ where $pq = c$ and $p + q = b$.
$x^2 + 5x + 6 = (x + 2)(x + 3) = 0 \Rightarrow x = -2$ or $x = -3$

Non-monic ($a \neq 1$) — AC method:
1. Multiply $a \times c$.
2. Find two numbers that multiply to $ac$ and add to $b$.
3. Split the middle term and factorise by grouping.

Example: $2x^2 + 7x + 3$
$a \times c = 6$. Need $p + q = 7$, $pq = 6$: use 6 and 1.
$= 2x^2 + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)$
$x = -\tfrac{1}{2}$ or $x = -3$

### Method 2 — QUADRATIC FORMULA (always works)

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

The DISCRIMINANT $\Delta = b^2 - 4ac$ tells you how many real solutions:

| $\Delta$ | Solutions |
|-----------|----------|
| $\Delta > 0$ | Two distinct real solutions |
| $\Delta = 0$ | One repeated (double) root: $x = -b/(2a)$ |
| $\Delta < 0$ | No real solutions (complex — VCE Methods) |

### Method 3 — COMPLETING THE SQUARE

Transform $x^2 + bx + c$ into $(x + h)^2 + k = 0$ form.

Step-by-step:
$x^2 + 6x + 5 = 0$
$(x^2 + 6x + 9) - 9 + 5 = 0$
$(x + 3)^2 - 4 = 0$
$(x + 3)^2 = 4$
$x + 3 = \pm 2$
$x = -1$ or $x = -5$

### Strategy Guide
• Try factorising first (fastest when it works).
• Use completing the square when you need vertex form for a parabola.
• Use the formula when coefficients are awkward or factorising fails.

### Connection to the Quadratic Formula
Completing the square on $ax^2 + bx + c = 0$ DERIVES the quadratic formula — you can derive it yourself as a proof exercise!

### Australian context
Quadratic equations model the trajectory of AFL footballs ($h = -4.9t^2 + vt + h_0$), the profit-maximisation problems in VCE Business Management, and the lens equations in optics used in Australian-built telescopes.

Quadratic Solving Formulas

  • Quadratic formula: x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}
  • Discriminant: \Delta = b^2 - 4ac
  • Completing the square: x^2 + bx = \left(x + \frac{b}{2}\right)^2 - \frac{b^2}{4}
  • Sum and product of roots: x_1 + x_2 = -\frac{b}{a}, \quad x_1 x_2 = \frac{c}{a}

Watch out

  • Sign in discriminant: It's $b^2 - 4ac$, not $b^2 + 4ac$. Get the sign wrong and everything fails.
  • Both solutions (±): The ± gives TWO solutions. Write BOTH unless the context rules one out.
  • Δ = 0 means ONE root: A double root: $x = -b/(2a)$. Not zero solutions — one "touching" solution.

Worked example (Medium)

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

  1. Identify. $a = 2$, $b = -5$, $c = 2$.
  2. Discriminant. $\Delta = (-5)^2 - 4(2)(2) = 25 - 16 = 9 > 0$ (two solutions).
  3. Formula. $x = \dfrac{5 \pm \sqrt{9}}{4} = \dfrac{5 \pm 3}{4}$
  4. Solutions. $x = \dfrac{8}{4} = 2$ or $x = \dfrac{2}{4} = \dfrac{1}{2}$

Answer: $x = 2$ or $x = \dfrac{1}{2}$.

Worked example (Hard)

Find the values of $k$ for which $x^2 - kx + (k+3) = 0$ has two distinct real solutions.

  1. Condition. Two distinct real solutions when $\Delta > 0$.
  2. Discriminant. $\Delta = (-k)^2 - 4(1)(k+3) = k^2 - 4k - 12$
  3. Inequality. $k^2 - 4k - 12 > 0$
  4. Factorise. $(k-6)(k+2) > 0$
  5. Sign analysis. Positive when $k > 6$ or $k < -2$.

Answer: $k > 6$ or $k < -2$.

Common mistakes

  • Forgetting the ± and writing only one solution from the formula The $\pm$ in the quadratic formula gives TWO solutions. Omitting it halves the answer. Fix: Write both: $x_1 = \frac{-b + \sqrt{\Delta}}{2a}$ and $x_2 = \frac{-b - \sqrt{\Delta}}{2a}$.
  • Calculating $(-b)^2$ incorrectly when $b$ is negative If $b = -5$, then $-b = 5$ and $(-b)^2 = 25$. But in the discriminant, it is $b^2 = (-5)^2 = 25$ — the same. Errors arise when students write $b^2 = -25$. Fix: Always square $b$ first (squaring removes the sign): $(-5)^2 = 25$, not $-25$.
  • Trying to factorise when the discriminant is not a perfect square If $\Delta$ is not a perfect square (e.g. $\Delta = 5$), the roots are irrational and cannot be expressed as simple fractions. Factorising won't give neat integers. Fix: Check $\Delta$ first. If not a perfect square integer, use the formula or leave in surd form.

Practice questions

  1. The discriminant of $ax^2 + bx + c$ is:

    1. $a^2 - 4bc$
    2. $b^2 - 4ac$
    3. $b^2 + 4ac$
    4. $4ac - b^2$

    (1 marks)

  2. Solve $x^2 - 6x + 5 = 0$ by factorising.

    (2 marks)

  3. Solve $3x^2 + 8x - 3 = 0$ using the quadratic formula.

    (3 marks)

  4. Complete the square to solve $x^2 + 4x - 1 = 0$. Leave answer in surd form.

    (3 marks)

  5. For what value(s) of $k$ does $x^2 + 2kx + 9 = 0$ have exactly one solution?

    (3 marks)

Going Deeper — Sum and Product of Roots

For $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$:

$$\alpha + \beta = -\frac{b}{a} \qquad \alpha \beta = \frac{c}{a}$$

These are called VIETA'S FORMULAS and can save time in certain problems.

Example: For $2x^2 - 5x + 2 = 0$ (roots 2 and 1/2):
Sum $= 2 + 1/2 = 5/2 = -(-5)/2 = b/a$ ... wait: $-b/a = -(-5)/2 = 5/2$ ✓
Product $= 2 \times 1/2 = 1 = c/a = 2/2 = 1$ ✓

Applications:
• Find an equation from roots: if roots are 3 and -4, then $x^2 - (3-4)x + (3)(-4) = 0 \Rightarrow x^2 + x - 12 = 0$.
• Find $\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$ without finding individual roots.

These formulas are heavily used in VCE Specialist Mathematics and competitions.

Summary

  • Three methods: factorising (fastest), quadratic formula (always works), completing the square.
  • Formula: $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
  • $\Delta > 0$: two distinct real roots; $\Delta = 0$: one double root; $\Delta < 0$: no real roots.
  • Non-monic factorising: AC method — multiply $a \times c$, split middle term, group.
  • Completing the square: adds and subtracts $(b/2)^2$ to form a perfect square trinomial.

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