Quadratic Equations
Every question with a fully worked solution.
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Theory
A quadratic equation has a squared pronumeral, such as \(3x^2 + 5 = 53\). When there is no \(x\) term, you solve it by getting \(x^2\) on its own and then taking the square root, which usually gives two answers, \(\pm\). This Year 9 maths page covers when there are two, one or no solutions, exact answers and rounded decimals, writing a quadratic in standard form, and quadratics that come from areas and Pythagoras' theorem.
A quadratic equation has a squared pronumeral as its highest power. Its standard form is \(ax^2 + bx + c = 0\), where \(a \ne 0\). Any quadratic equation can be rearranged into this form by expanding brackets and moving every term to one side.
On most of this page there is no \(x\) term, 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 square root such as \(\sqrt{7}\), or as a decimal rounded as asked.
You can also solve graphically or numerically. The solutions of \(x^2 = k\) are the \(x\)-values where the parabola \(y = x^2\) meets the line \(y = k\). Or try values on a calculator: \(2.2^2 = 4.84\) and \(2.3^2 = 5.29\), so the positive solution of \(x^2 = 5\) lies between \(2.2\) and \(2.3\) (it is \(2.236\ldots\)).
| 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 \(\pm\).
- Leave the root exact, 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.
- A minus sign in front of \(x^2\): add the \(x^2\) term to both sides, or multiply through by \(-1\), so that \(x^2\) is positive.
- Standard form: expand any brackets, then move every term to one side so the other side is \(0\).
- In context: lengths, areas, 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 \(5(-5)^2 - 7 = 118\).
| \(x^2 + 4\) | \(=\) | \(36\) |
| \(x^2\) | \(=\) | \(32\) |
| \(x\) | \(=\) | \(\mathbf{\pm\sqrt{32}}\) |
That is the exact answer. On a calculator \(\sqrt{32} = 5.656\ldots\), so
| \(x\) | \(\approx\) | \(\mathbf{\pm 5.66}\) |
(i) \(2x^2 + 9 = 1\)
(ii) \(6x^2 - 11 = x^2 + 34\)
(iii) \(7 - \dfrac{x^2}{2} = 7\)
(i) Isolate \(x^2\) first, then look at its sign.
| \(2x^2\) | \(=\) | \(-8\) |
| \(x^2\) | \(=\) | \(-4\) |
No real number squares to a negative, so there is no solution.
(ii) Collect the \(x^2\) terms on the left and the numbers on the right.
| \(5x^2\) | \(=\) | \(45\) |
| \(x^2\) | \(=\) | \(9\) |
| \(x\) | \(=\) | \(\mathbf{\pm 3}\) |
(iii) Subtract \(7\) from both sides.
| \(-\dfrac{x^2}{2}\) | \(=\) | \(0\) |
| \(x^2\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\mathbf{0}\) |
Zero has only one square root, so (iii) has exactly one solution.
The diagonal is the hypotenuse of a right-angled triangle with two sides \(s\).
| \(s^2 + s^2\) | \(=\) | \(14^2\) |
| \(2s^2\) | \(=\) | \(196\) |
| \(s^2\) | \(=\) | \(98\) |
| \(s\) | \(=\) | \(\pm\sqrt{98}\) |
A side cannot be negative, so \(s = \mathbf{\sqrt{98}}\) 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 give an exact answer?
Leave the square root in the answer: \(x^2 = 11\) gives \(x = \pm\sqrt11\). If a decimal is asked for, round only at the end: \(\pm\sqrt11 \approx \pm 3.32\) to two decimal places.
What is the standard form of a quadratic equation?
It is \(ax^2 + bx + c = 0\) with \(a \ne 0\). Expand any brackets and move every term to one side. For example \(x(x + 3) = 2x + 7\) becomes \(x^2 + x - 7 = 0\).
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.