Simple Quadratic Equations
A quadratic equation contains a squared pronumeral. This NSW Year 8 maths subtopic (NESA outcome MA4‑EQU‑C‑01) covers equations of the form \(ax^2 = c\): get \(x^2\) on its own, then take the square root of both sides, remembering that \(x^2 = 49\) gives \(x = 7\) or \(x = -7\).
You will solve equations such as \(3x^2 + 8 = 35\) and \(\frac{x^2}{3} = 12\), explain why \(3x^2 = -27\) has no solution, and give answers exactly as a surd, like \(\pm\sqrt{13}\), or as a rounded decimal. You will also use formulas, finding a radius from \(A = \pi r^2\) or the time a dropped stone takes to fall, and reject the negative root when a length or time cannot be negative. It is part of Year 8 Equations.
Theory
A simple quadratic equation has a squared pronumeral. Get \(x^2\) on its own, then take the square root of both sides, and write both roots: \(x^2 - 8 = 92\) gives \(x^2 = 100\), so \(x = \pm 10\). This Year 8 maths page shows how.
Why two answers. \(8 \times 8 = 64\) and \((-8) \times (-8) = 64\), so \(x^2 = 64\) has two solutions, \(x = 8\) and \(x = -8\), written together as \(x = \pm 8\).
The method. Undo \(+\) or \(-\) first, then \(\times\) or \(\div\), until \(x^2\) is alone. Then square root both sides: when \(c > 0\), \(x^2 = c\) gives \(x = \pm\sqrt{c}\).
Decimals work the same way. \(0.6 \times 0.6 = 0.36\), so \(x^2 = 0.36\) gives \(x = \pm 0.6\).
Exact form or a decimal. If \(c\) is not a square number, \(\sqrt{c}\) is a surd, the exact answer. A calculator gives a decimal approximation, rounded as the question asks.
No solution. A square is never negative, so \(x^2 = -16\) has no solution. (\(x^2 = 0\) has the single solution \(x = 0\).)
A squared bracket. In \((2x + 1)^2 = 25\), square root first: \(2x + 1 = \pm 5\). Then solve the two equations \(2x + 1 = 5\) and \(2x + 1 = -5\) separately.
Formulas, lengths and times. Substitute what you know, get the squared pronumeral alone, then square root. A length, radius, speed or time cannot be negative, so reject the negative root and say why.
Get x² alone, then square root:
| Equation | Get \(x^2\) alone | Solutions |
|---|---|---|
| \(x^2 = 196\) | already alone | \(x = \pm 14\) |
| \(x^2 + 4 = 68\) | \(x^2 = 64\) | \(x = \pm 8\) |
| \(5x^2 = 80\) | \(x^2 = 16\) | \(x = \pm 4\) |
| \(\dfrac{x^2}{4} = 25\) | \(x^2 = 100\) | \(x = \pm 10\) |
| \(2x^2 = 42\) | \(x^2 = 21\) | \(x = \pm\sqrt{21} \approx \pm 4.58\) |
| \(3x^2 = -48\) | \(x^2 = -16\) | no solution |
How to solve ax² + b = c
- Undo the added or subtracted number on both sides.
- Undo the multiplying or dividing, so \(x^2\) is alone.
- Square root both sides and write \(\pm\). If \(x^2\) is negative, there is no solution.
- Round if asked, and reject a negative root only when the context says so.
How to solve a squared bracket
- Square root both sides: the bracket equals \(+\) or \(-\) the root.
- Solve the two linear equations separately.
| \(x^2 - 8 + 8\) | \(=\) | \(92 + 8\) |
| \(x^2\) | \(=\) | \(100\) |
| \(x\) | \(=\) | \(\pm\sqrt{100} = \pm 10\) |
\(x = \pm 10\). Check: \(10^2 - 8 = 92\) and \((-10)^2 - 8 = 92\).
| \(2x^2\) | \(=\) | \(71 + 1 = 72\) |
| \(x^2\) | \(=\) | \(\dfrac{72}{2} = 36\) |
| \(x\) | \(=\) | \(\pm\sqrt{36} = \pm 6\) |
\(x = \pm 6\)
(ii) Solve \(4x^2 = -36\), or explain why it has no solution.
| \(x^2\) | \(=\) | \(\dfrac{55}{5} = 11\) |
| \(x\) | \(=\) | \(\pm\sqrt{11} \approx \pm 3.32\) |
| \(x^2\) | \(=\) | \(\dfrac{-36}{4} = -9\) |
(i) \(11\) is not a square number, so \(x = \pm\sqrt{11}\) exactly, or \(x \approx \pm 3.32\).
(ii) No solution: no number squared gives \(-9\).
| \(2x + 1\) | \(=\) | \(\pm\sqrt{25} = \pm 5\) |
| \(2x + 1 = 5\) | \(\text{ or }\) | \(2x + 1 = -5\) |
| \(2x = 4\) | \(\text{ or }\) | \(2x = -6\) |
| \(x = 2\) | \(\text{ or }\) | \(x = -3\) |
\(x = 2\) or \(x = -3\). Check: \(5^2 = 25\) and \((-5)^2 = 25\).
| \(\pi r^2\) | \(=\) | \(1200\) |
| \(r^2\) | \(=\) | \(\dfrac{1200}{\pi} \approx 381.97\) |
| \(r\) | \(\approx\) | \(\pm 19.54\) |
A radius cannot be negative, so reject \(-19.54\). The radius is about \(19.5\) cm.
| \(\dfrac{v^2}{150}\) | \(=\) | \(54\) |
| \(v^2\) | \(=\) | \(54 \times 150 = 8100\) |
| \(v\) | \(=\) | \(\pm\sqrt{8100} = \pm 90\) |
A speed cannot be negative, so reject \(-90\). The car was going \(90\) km/h.
Volume \(=\) base area \(\times\) height, and the base area is \(x^2\).
| \(10x^2\) | \(=\) | \(2000\) |
| \(x^2\) | \(=\) | \(200\) |
| \(x\) | \(\approx\) | \(\pm 14.14\) |
A length cannot be negative, so each side of the base is about \(14.1\) cm.
Common pitfalls
Frequently asked questions
Why does x squared = 25 have two solutions?
Because \(5^2 = 25\) and \((-5)^2 = 25\), so \(x = \pm 5\).
What does ± mean?
Plus or minus: \(x = \pm 3\) means \(x = 3\) or \(x = -3\).
When is there no solution?
When the square equals a negative number, such as \(x^2 = -4\): no number squared is negative.
What is an exact answer?
An answer left as a surd, such as \(\sqrt{7}\), instead of a rounded decimal.
Why reject the negative root sometimes?
A length, radius, speed or time cannot be negative, so in those problems only the positive root makes sense.