Quadratic Equations
A quadratic equation contains the square of the pronumeral. This NSW Year 9 Core maths subtopic solves the simplest kind, with no middle term, such as \(x^2 = 36\) or \(4x^2 - 9 = 27\). It goes a little beyond the Core outcomes and comes from NESA outcome MA5‑EQU‑P‑01.
You will isolate the squared term by clearing constants, coefficients and fractions first, then take the square root and give both solutions, so \(x^2 = 25\) gives \(x = \pm 5\), not just \(5\). You will round solutions to one decimal place, leave answers in simplest surd form, such as \(x = \pm 2\sqrt{5}\), and reject the negative solution when it makes no sense, as for the side of a square garden bed with an area of \(121~\text{m}^2\). It is part of Year 9 Core Equations.
Theory
A quadratic equation has a squared pronumeral, such as \(x^2 = 25\). This Year 9 maths page shows how to get the square on its own, take the square root of both sides and give both solutions. It also covers rounding, surd answers and practical problems. Quadratic equations come up if you are heading for the Advanced path.
A quadratic equation has the pronumeral squared, such as \(x^2 = 25\) or \(2x^2 + 3 = 21\). On this page there is no \(x\) term on its own.
Both \(5\) and \(-5\) square to \(25\). So \(x^2 = 25\) has two solutions: \(x = 5\) or \(x = -5\). This is written \(x = \pm 5\), read "plus or minus \(5\)".
First get \(x^2\) on its own, undoing the steps in reverse order. Then take the square root of both sides.
If the number is not a perfect square, the answer is a surd such as \(\sqrt{11}\). Leave it as a surd for an exact answer, or round it.
The sign \(\sqrt{\phantom{9}}\) on its own means the positive root. You add the \(\pm\) yourself when solving.
Once the square is on its own:
| Equation | Isolate the square | Solutions |
|---|---|---|
| \(x^2 = 9\) | already isolated | \(x = \pm 3\) |
| \(4x^2 = 9\) | \(x^2 = \dfrac{9}{4}\) | \(x = \pm \dfrac{3}{2}\) |
| \(x^2 - 5 = 4\) | \(x^2 = 9\) | \(x = \pm 3\) |
| \(\dfrac{x^2}{4} = 9\) | \(x^2 = 36\) | \(x = \pm 6\) |
How to solve a quadratic equation like \(ax^2 + c = k\)
- Isolate \(x^2\): undo any adding, then any multiplying or dividing.
- Take the square root of both sides and write \(\pm\).
- Simplify a surd, or round as the question asks.
- Check both answers, and reject a negative one in a practical problem.
(i) \(x^2 = 64\)
(ii) \(x^2 = 0.36\).
The square is already on its own, so take the square root of both sides. Write \(\pm\) for the two answers.
(i)
| \(x\) | \(=\) | \(\pm\sqrt{64}\) |
| \(\) | \(=\) | \(\boldsymbol{\pm 8}\) |
Check both: \(8^2 = 64\) and \((-8)^2 = 64\) \(\checkmark\)
(ii)
| \(x\) | \(=\) | \(\pm\sqrt{0.36}\) |
| \(\) | \(=\) | \(\boldsymbol{\pm 0.6}\) |
\(0.6 \times 0.6 = 0.36\). Count the decimal places: two in the square, one in the root.
(i) \(3x^2 = 108\)
(ii) \(\dfrac{x^2}{6} = 24\).
(i) Divide both sides by \(3\), then take the root.
| \(x^2\) | \(=\) | \(\dfrac{108}{3}\) |
| \(x^2\) | \(=\) | \(36\) |
| \(x\) | \(=\) | \(\boldsymbol{\pm 6}\) |
(ii) Multiply both sides by \(6\), then take the root.
| \(x^2\) | \(=\) | \(24 \times 6\) |
| \(x^2\) | \(=\) | \(144\) |
| \(x\) | \(=\) | \(\boldsymbol{\pm 12}\) |
(i) \(2x^2 + 7 = 39\)
(ii) \(\dfrac{x^2 + 4}{3} = 5\), correct to one decimal place.
Undo the steps in reverse order until \(x^2\) is on its own.
(i)
| \(2x^2\) | \(=\) | \(32\) |
| \(x^2\) | \(=\) | \(16\) |
| \(x\) | \(=\) | \(\boldsymbol{\pm 4}\) |
(ii) Clear the fraction first, then subtract \(4\).
| \(x^2 + 4\) | \(=\) | \(15\) |
| \(x^2\) | \(=\) | \(11\) |
| \(x\) | \(=\) | \(\pm\sqrt{11}\) |
| \(\) | \(=\) | \(\pm 3.316\ldots\) |
| \(\) | \(\approx\) | \(\boldsymbol{\pm 3.3}\) |
(i) Solve \(x^2 = 20\) in simplest surd form.
(ii) A square tile has an area of \(50 \text{ cm}^2\). Find its side length, correct to two decimal places.
(i) \(20\) is not a perfect square. Split off its largest square factor, \(4\).
| \(x\) | \(=\) | \(\pm\sqrt{20}\) |
| \(\) | \(=\) | \(\pm\sqrt{4 \times 5}\) |
| \(\) | \(=\) | \(\boldsymbol{\pm 2\sqrt{5}}\) |
(ii) Let the side be \(s\) cm. Then \(s^2 = 50\).
| \(s\) | \(=\) | \(\pm\sqrt{50}\) |
| \(\) | \(=\) | \(\pm 7.071\ldots\) |
A length cannot be negative, so reject \(-7.07\). The side is 7.07 cm.
Common pitfalls
Frequently asked questions
Why does a quadratic equation have two answers?
Because a positive number and its negative have the same square. Both \(5\) and \(-5\) square to \(25\), so \(x^2 = 25\) gives \(x = \pm 5\).
How do you solve x squared equals a number?
Take the square root of both sides and write \(\pm\). If \(x^2 = 64\), then \(x = \pm 8\).
What does the plus or minus sign mean?
It is a short way of writing two answers. \(x = \pm 6\) means \(x = 6\) or \(x = -6\).
How do you simplify a surd answer?
Split the number under the root into a square number times another number, using the largest square you can: \(\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}\).
When do you reject the negative answer?
When the pronumeral stands for something that cannot be negative, such as a length, a time or a number of people. Give only the positive answer and say why.