Quadratic Equations
Master quadratic equations for Queensland Year 11 Mathematical Methods (QCAA). A quadratic equation sets a squared expression equal to zero, and usually has two solutions.
You will learn to solve by factorising and the null factor law, take square roots to handle bracket-squared forms, and turn worded problems into equations while rejecting answers that make no sense — a core skill behind every parabola you sketch.
Theory
In Year 11 Mathematical Methods (QCAA, Unit 1), a quadratic equation is solved by making one side zero, factorising, and applying the null factor law: if a product is zero, at least one factor is zero. This page covers the null factor law, solving monic and non-monic quadratics by factorisation, solving \((x-h)^2=k\) by square roots, and worded problems that reduce to a quadratic.
A quadratic equation can be written \(ax^2+bx+c=0\) with \(a\ne0\). Its solutions (also called roots) are the \(x\)-values that make it true — the \(x\)-intercepts of the matching parabola.
The null factor law states that if \(A\times B=0\) then \(A=0\) or \(B=0\). So once a quadratic is written as a product equal to zero, you set each factor to zero and solve the simple linear equations.
When a quadratic is a perfect square equal to a number, \((x-h)^2=k\) with \(k\ge0\), you may take square roots of both sides, remembering the \(\pm\).
The two tools for this subtopic:
How to solve a quadratic equation
- Rearrange so one side is \(0\) (collect all terms on the left).
- Factorise the quadratic, or if it is \((x-h)^2=k\) take square roots of both sides.
- Solve each factor equal to zero, then state both roots (and reject any that do not fit a worded context).
The product is already zero, so apply the null factor law — set each factor to zero:
| \(x-3\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(3\) |
and the other factor:
| \(x+4\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(-4\) |
\(x=3\) or \(x=-4\).
Make one side zero:
| \(x^2+3x-10\) | \(=\) | \(0\) |
Factorise the monic trinomial — two numbers multiplying to \(-10\), adding to \(3\) are \(5\) and \(-2\):
| \((x+5)(x-2)\) | \(=\) | \(0\) |
Apply the null factor law:
| \(x+5\) | \(=\) | \(0\ \Rightarrow\ x=-5\) |
| \(x-2\) | \(=\) | \(0\ \Rightarrow\ x=2\) |
\(x=-5\) or \(x=2\).
Take the square root of both sides, keeping \(\pm\):
| \(x-2\) | \(=\) | \(\pm\sqrt{25}\) |
| \(x-2\) | \(=\) | \(\pm5\) |
Solve the two cases:
| \(x\) | \(=\) | \(2+5=7\) |
| \(x\) | \(=\) | \(2-5=-3\) |
\(x=7\) or \(x=-3\).
Let the width be \(w\) metres, so the length is \(w+3\). Area gives an equation:
| \(w(w+3)\) | \(=\) | \(40\) |
Expand and make one side zero:
| \(w^2+3w\) | \(=\) | \(40\) |
| \(w^2+3w-40\) | \(=\) | \(0\) |
Factorise — two numbers multiplying to \(-40\), adding to \(3\) are \(8\) and \(-5\):
| \((w+8)(w-5)\) | \(=\) | \(0\) |
| \(w\) | \(=\) | \(-8\ \text{or}\ 5\) |
A width cannot be negative, so reject \(w=-8\).
| \(w\) | \(=\) | \(5\) |
The width is \(5\ \text{m}\) (and the length \(8\ \text{m}\)).
Common pitfalls
Frequently asked questions
What is the null factor law?
If a product of factors equals zero, then at least one factor is zero. So from \((x-a)(x-b)=0\) you get \(x=a\) or \(x=b\).
Why must one side be zero before factorising?
Because the null factor law only applies to a product equal to zero. If the right side is not zero, a factor being that number tells you nothing.
How do I solve \((x-h)^2=k\)?
Take the square root of both sides to get \(x-h=\pm\sqrt{k}\), then solve the two cases \(x=h+\sqrt{k}\) and \(x=h-\sqrt{k}\).
How many solutions does a quadratic have?
Usually two. They coincide when the quadratic is a perfect square, and there are no real solutions when a square would have to equal a negative number.
Do the solutions relate to the graph?
Yes — the real solutions are exactly the \(x\)-intercepts of the parabola \(y=ax^2+bx+c\).
Why reject a negative answer in a word problem?
Lengths, times and counts cannot be negative, so a negative root is not a valid answer even though it solves the equation.