Solving Quadratic Equations Using Factorisation
Every question with a fully worked solution.
- Solving Quadratic Equations Using Factorisation - Video - Solving Quadratic Equations by Factoring - Basic Examples Watch
Theory
A quadratic equation such as \(x^2 - 5x + 6 = 0\) usually has two solutions. This Year 10 maths page shows how to solve it by factorising and using the null factor law, how to rearrange an equation so that one side is zero, how to handle fractions and letters, and how to set up and solve quadratic equations from real problems.
A quadratic equation has \(x^2\) as its highest power. Its standard form is \(x^2 + bx + c = 0\), with every term on one side and zero on the other.
The null factor law: if two numbers multiply to give zero, at least one of them is zero. So if \((x - 2)(x - 5) = 0\), then \(x - 2 = 0\) or \(x - 5 = 0\).
The values of \(x\) that make the equation true are its solutions (or roots). A quadratic equation can have two solutions, one solution (when it is a perfect square), or none.
In a problem, a solution may not make sense, such as a negative length or time. Reject it and say why.
The null factor law:
So \((x - p)(x - q) = 0\) has solutions \(x = p\) and \(x = q\). Notice the signs change: \((x + 6) = 0\) gives \(x = -6\).
How to solve a quadratic equation by factorising
- Expand any brackets if the right side is not zero.
- Rearrange so that one side is \(0\), with the \(x^2\) term positive.
- Divide through by any common number factor (never by \(x\)).
- Factorise the quadratic.
- Use the null factor law: set each factor equal to \(0\) and solve.
- Check each solution in the original equation. In a problem, reject any that do not make sense.
Other question types
- Fractions: note the value that makes a denominator zero, multiply every term by the denominator, then solve the quadratic (Example 4).
- A common bracket on both sides: in \((x + 1)^2 = 3x(x + 1)\), bring everything to one side and take out \((x + 1)\): \((x + 1)(1 - 2x) = 0\), so \(x = -1\) or \(x = \dfrac{1}{2}\). Dividing by \(x + 1\) loses \(x = -1\).
- Letters as coefficients: treat the letter as a number. \(x^2 - ax - 6a^2 = (x - 3a)(x + 2a)\), so \(x = 3a\) or \(x = -2a\).
- A parameter \(k\): factorise to find the solutions in terms of \(k\), then apply the condition given, such as two solutions that differ by a given amount. Remember both orders, such as \(k - 2\) and \(2 - k\).
How to set up a quadratic equation from a problem
- Choose a pronumeral for one unknown and write the others in terms of it.
- Use the fact in the question (an area, Pythagoras' theorem, a product, a formula) to write an equation.
- Solve, then answer the question in words with units.
(i) \((x + 4)(x - 7) = 0\)
(ii) \(5x^2 - 20x = 0\)
(i) One of the brackets must be zero.
| \(x + 4 = 0\) | \(\text{or}\) | \(x - 7 = 0\) |
| \(\boldsymbol{x = -4}\) | \(\text{or}\) | \(\boldsymbol{x = 7}\) |
(ii) Factorise with the HCF \(5x\). Do not divide both sides by \(x\): that loses the solution \(x = 0\).
| \(5x(x - 4)\) | \(=\) | \(0\) |
| \(5x = 0\) | \(\text{or}\) | \(x - 4 = 0\) |
| \(\boldsymbol{x = 0}\) | \(\text{or}\) | \(\boldsymbol{x = 4}\) |
(i) \(x^2 = 7x - 10\)
(ii) \((x - 2)(x + 5) = 8\)
(i) Move every term to the left side, then factorise.
| \(x^2 - 7x + 10\) | \(=\) | \(0\) |
| \((x - 2)(x - 5)\) | \(=\) | \(0\) |
| \(\boldsymbol{x = 2}\) | \(\text{or}\) | \(\boldsymbol{x = 5}\) |
(ii) The right side is \(8\), not \(0\), so the null factor law cannot be used yet. Expand and subtract \(8\).
| \(x^2 + 3x - 10\) | \(=\) | \(8\) |
| \(x^2 + 3x - 18\) | \(=\) | \(0\) |
| \((x + 6)(x - 3)\) | \(=\) | \(0\) |
| \(\boldsymbol{x = -6}\) | \(\text{or}\) | \(\boldsymbol{x = 3}\) |
Check \(x = 3\): \((3 - 2)(3 + 5) = 1 \times 8 = 8\). ✓
(i) \(3x^2 - 3x - 36 = 0\)
(ii) \(x^2 + 14x + 49 = 0\)
(i) Divide both sides by \(3\) (it is not zero), then factorise.
| \(x^2 - x - 12\) | \(=\) | \(0\) |
| \((x - 4)(x + 3)\) | \(=\) | \(0\) |
| \(\boldsymbol{x = -3}\) | \(\text{or}\) | \(\boldsymbol{x = 4}\) |
(ii) This is a perfect square, so both brackets give the same solution.
| \((x + 7)^2\) | \(=\) | \(0\) |
| \(x + 7\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\boldsymbol{-7}\) |
A perfect square gives only one solution.
(i) \(x - \dfrac{10}{x} = 3\)
(ii) \(\dfrac{x}{3} + \dfrac{4}{x} = \dfrac{7}{3}\)
(i) \(x \neq 0\). Multiply every term by \(x\), then make one side zero.
| \(x^2 - 10\) | \(=\) | \(3x\) |
| \(x^2 - 3x - 10\) | \(=\) | \(0\) |
| \((x - 5)(x + 2)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\boldsymbol{5} \text{ or } \boldsymbol{x = -2}\) |
Neither value is \(0\), so both are solutions.
(ii) \(x \neq 0\). Multiply every term by \(3x\), then make one side zero.
| \(x^2 + 12\) | \(=\) | \(7x\) |
| \(x^2 - 7x + 12\) | \(=\) | \(0\) |
| \((x - 3)(x - 4)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\boldsymbol{3} \text{ or } \boldsymbol{x = 4}\) |
Check: \(\dfrac{3}{3} + \dfrac{4}{3} = \dfrac{7}{3}\) and \(\dfrac{4}{3} + \dfrac{4}{4} = \dfrac{7}{3}\). ✓
Common pitfalls
Frequently asked questions
What is the null factor law?
If two numbers multiply to give zero, at least one of them must be zero. It lets you solve \((x - p)(x - q) = 0\) by setting each bracket equal to zero.
Why must one side be zero?
The null factor law only works for a product equal to zero. If \((x - 1)(x + 2) = 10\), the brackets could be \(2\) and \(5\), or many other pairs, so you must rearrange first.
How many solutions can a quadratic equation have?
Two, one or none. A perfect square such as \((x - 3)^2 = 0\) has only one solution, \(x = 3\).
Why do some answers get rejected?
In a real problem a solution can be impossible, such as a negative length or a negative time. Both values solve the equation, but only one answers the question.
How do I check my solutions?
Substitute each solution into the original equation. Both sides should give the same value.