Quadratic Equations
A quadratic equation of the form \(ax^2 = c\) is solved by making \(x^2\) the subject and taking the square root, which gives two answers, \(\pm\), when \(c\) is positive. This NSW Year 9 Advanced maths subtopic extends that Stage 4 skill (NESA outcome MA4‑EQU‑C‑01).
You will solve equations such as \(3k^2 = 192\) and \(5 - \frac{d^2}{3} = -2\), give answers as exact surds like \(\pm 3\sqrt{5}\) or rounded, and recognise when there is no real solution. You will solve \((x - 3)^2 = 49\), find the width of a rectangle from its diagonal and explain why only one solution fits, and find the values of \(k\) that give two, one or no solutions. It is part of Year 9 Advanced Equations.
Theory
A quadratic equation such as \(3x^2 + 5 = 53\) is solved by getting \(x^2\) on its own and then taking the square root, which gives two answers, \(\pm\). This Year 9 Advanced maths page covers when there are two, one or no solutions, answers in simplest surd form and as decimals, squared brackets, and quadratics that come from Pythagoras' theorem. The surd answers are useful if you are heading for Advanced maths.
A quadratic equation has a squared pronumeral as its highest power. On this page the pronumeral appears only as \(x^2\) (or inside one squared bracket), so the equation can be written as \(x^2 = k\).
Both \(4^2\) and \((-4)^2\) equal \(16\), so \(x^2 = 16\) has two solutions, \(x = \pm 4\), read "plus or minus \(4\)".
A square is never negative. So \(x^2 = k\) has two solutions if \(k > 0\), one solution (\(x = 0\)) if \(k = 0\), and no real solution if \(k \lt 0\).
Solve by undoing operations in reverse order until \(x^2\) is alone, and then take the square root. The root of a sum is not the sum of the roots: \(\sqrt{36 + 64} = 10\), but \(\sqrt{36} + \sqrt{64} = 14\).
When the root is not a whole number, give it either exactly as a surd in simplest form, or as a decimal rounded as asked. A surd is in simplest form when no square number other than \(1\) divides the number under the root.
Simplifying surds with the largest square factor:
| Value of \(k\) in \(x^2 = k\) | Solutions |
|---|---|
| \(k > 0\) | two: \(\pm\sqrt{k}\) |
| \(k = 0\) | one: \(0\) |
| \(k \lt 0\) | none (no real number squares to a negative) |
How to solve \(ax^2 + b = c\) and similar equations
- Isolate \(x^2\): undo additions and subtractions, then multiplication and division (clear any fraction first if that is easier).
- Decide how many solutions: look at the sign of the number now equal to \(x^2\).
- Take the square root of both sides and write \(\pm\).
- Simplify the surd, or round as asked. Keep the exact value in the calculator until the end.
- Check both answers in the original equation.
Variations
- \(x^2\) on both sides: collect the \(x^2\) terms on one side first, exactly as with \(x\) terms in a linear equation.
- Squared bracket: root both sides straight away, then solve the two linear equations \(x + b = +\sqrt{k}\) and \(x + b = -\sqrt{k}\).
- A constant in the equation: to find which values of a constant give two, one or no solutions, make \(x^2\) the subject and ask when the right side is positive, zero or negative.
- In context: lengths, times and speeds must be positive. State that you reject the negative root and why.
Undo the \(-7\), then the \(\times 5\), and only then take the square root.
| \(5x^2\) | \(=\) | \(125\) |
| \(x^2\) | \(=\) | \(25\) |
| \(x\) | \(=\) | \(\pm\sqrt{25}\) |
| \(\) | \(=\) | \(\mathbf{\pm 5}\) |
Check both: \(5(5)^2 - 7 = 118\) and \(5(-5)^2 - 7 = 118\).
| \(x^2 + 4\) | \(=\) | \(36\) |
| \(x^2\) | \(=\) | \(32\) |
| \(x\) | \(=\) | \(\pm\sqrt{32}\) |
The largest square factor of \(32\) is \(16\):
| \(\sqrt{32}\) | \(=\) | \(\sqrt{16 \times 2} = 4\sqrt{2}\) |
| \(x\) | \(=\) | \(\mathbf{\pm 4\sqrt{2}}\) |
| \(\) | \(\approx\) | \(\pm 5.66\) |
Use \(16\), not \(4\): \(\sqrt{4 \times 8} = 2\sqrt{8}\) is not yet in simplest form.
Take the square root of both sides first, keeping \(\pm\). Do not expand the bracket.
| \(x + 5\) | \(=\) | \(\pm\sqrt{12}\) |
| \(x + 5\) | \(=\) | \(\pm 2\sqrt{3}\) |
| \(x\) | \(=\) | \(\mathbf{-5 \pm 2\sqrt{3}}\) |
That is \(x \approx -1.54\) or \(x \approx -8.46\). Check: \((-1.536\ldots + 5)^2 = 3.464\ldots^2 = 12\).
The diagonal is the hypotenuse of a right-angled triangle with two sides \(s\).
| \(s^2 + s^2\) | \(=\) | \(14^2\) |
| \(2s^2\) | \(=\) | \(196\) |
| \(s^2\) | \(=\) | \(98\) |
| \(s\) | \(=\) | \(\pm\sqrt{98} = \pm 7\sqrt{2}\) |
A side cannot be negative, so \(s = \mathbf{7\sqrt{2}}\) cm \(\approx \mathbf{9.9}\) cm.
Common pitfalls
Frequently asked questions
Why does x squared = 25 have two answers?
Because \(5^2\) and \((-5)^2\) both equal \(25\). So \(x = 5\) or \(x = -5\), written \(x = \pm 5\).
When does a quadratic equation have no solution?
An equation of the form \(x^2 = k\) has no real solution when \(k\) is negative, because no real number squared gives a negative answer.
How do you write an answer in simplest surd form?
Find the largest square number that divides the number under the root, split the root and simplify the square part. For example \(\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}\).
How do you solve (x - 2) squared = 9?
Take the square root of both sides: \(x - 2 = 3\) or \(x - 2 = -3\). So \(x = 5\) or \(x = -1\). There is no need to expand the bracket.
When do you ignore the negative solution?
When the pronumeral stands for something that cannot be negative, such as a length, a time or a speed. Say that the negative root is rejected and why.