Binomial Products
Every question with a fully worked solution.
- Binomial Products - Video - Surds/Radicals 4 - Binomial Products Watch
Theory
To expand a binomial product with surds, multiply each term in the first bracket by each term in the second, then collect the whole numbers and the like surds. This Year 10A maths page covers perfect squares, the difference of two squares and conjugates, cubes, answers in the form \(a+b\sqrt{n}\) and finding unknowns by equating parts.
A binomial product is two brackets, each with two terms, multiplied together. Surds expand by the same rules as pronumerals: \((a+b)(c+d) = ac+ad+bc+bd\).
The difference is what happens next. A surd times itself is rational, \(\sqrt{3}\times\sqrt{3} = 3\), so some products become whole numbers. Collect the rational part and the surd part separately, and simplify every surd.
Two special products save time. A perfect square \((a\pm b)^2 = a^2\pm 2ab+b^2\) always has a middle term. The difference of two squares \((a+b)(a-b) = a^2-b^2\) has none. So \((5+\sqrt{2})(5-\sqrt{2}) = 25-2 = 23\) is rational. \(5+\sqrt{2}\) and \(5-\sqrt{2}\) are conjugates.
Answers are often asked for in the form \(a+b\sqrt{n}\). Two such expressions are equal only when their rational parts are equal and their surd parts are equal (for integers \(a, b\) and a surd \(\sqrt{n}\)). This lets you find unknowns.
| Product | Result |
|---|---|
| \((\sqrt{a})^2\) | \(a\) |
| \((\sqrt{a}+\sqrt{b})^2\) | \(a+b+2\sqrt{ab}\) |
| \((\sqrt{a}-\sqrt{b})^2\) | \(a+b-2\sqrt{ab}\) |
| \((\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})\) | \(a-b\) |
| \((p+\sqrt{b})(p-\sqrt{b})\) | \(p^2-b\) |
How to expand a binomial product with surds
- Check for a special product: a square \((a\pm b)^2\) or conjugates \((a+b)(a-b)\).
- Expand: four products, keeping track of signs. Multiply outside numbers together and inside numbers together.
- Simplify: \((\sqrt{a})^2 = a\), and simplify every surd, such as \(\sqrt{50} = 5\sqrt{2}\).
- Collect the rational terms, then each family of like surds.
- Write the answer in the form asked for, rational part first.
Other question types
- Pronumerals: \((\sqrt{x})^2 = x\) (not \(x^2\)), so \((\sqrt{x}+1)(\sqrt{x}-4) = x-3\sqrt{x}-4\).
- Cubes: square the bracket first, then multiply by it once more (Example 3).
- Substitution: to find \(x^2+kx+c\) when \(x = p+\sqrt{q}\), find \(x^2\) first, then substitute and collect. The surd parts often cancel.
- Unknown coefficients: expand, equate the rational parts and the surd parts, and solve the two equations, testing whole-number values if needed.
- Areas and costs: area \(= l\times w\) gives an exact answer in the form \(a+b\sqrt{n}\). Use a calculator only at the end, for a cost or a rounded length.
- Square roots of \(a+b\sqrt{n}\): spot or build the square it came from, then choose the positive root (Example 4).
(i) \((4\sqrt{3}-1)(2\sqrt{3}+5)\)
(ii) \((\sqrt{10}+\sqrt{6})(\sqrt{5}-\sqrt{3})\)
(i) Multiply each term of the first bracket by each term of the second.
| \(4\sqrt{3}\times 2\sqrt{3}\) | \(=\) | \(24\) |
| \(4\sqrt{3}\times 5\) | \(=\) | \(20\sqrt{3}\) |
| \(-1\times 2\sqrt{3}\) | \(=\) | \(-2\sqrt{3}\) |
| \(-1\times 5\) | \(=\) | \(-5\) |
| \(\text{sum}\) | \(=\) | \(24+20\sqrt{3}-2\sqrt{3}-5\) |
| \(\) | \(=\) | \(\mathbf{19+18\sqrt{3}}\) |
\(4\sqrt{3}\times 2\sqrt{3} = 8\times 3 = 24\). Collect the whole numbers and the surds separately.
(ii) Expand, then simplify each surd.
| \(\) | \(\) | \(\sqrt{50}-\sqrt{30}+\sqrt{30}-\sqrt{18}\) |
| \(\) | \(=\) | \(5\sqrt{2}-3\sqrt{2}\) |
| \(\) | \(=\) | \(\mathbf{2\sqrt{2}}\) |
The two \(\sqrt{30}\) terms cancel.
(i) \((3\sqrt{5}-2\sqrt{3})^2\)
(ii) \((4+\sqrt{7})(4-\sqrt{7})\)
(i) Use \((a-b)^2 = a^2-2ab+b^2\) with \(a = 3\sqrt{5}\) and \(b = 2\sqrt{3}\).
| \(a^2\) | \(=\) | \((3\sqrt{5})^2 = 45\) |
| \(2ab\) | \(=\) | \(2\times 3\sqrt{5}\times 2\sqrt{3} = 12\sqrt{15}\) |
| \(b^2\) | \(=\) | \((2\sqrt{3})^2 = 12\) |
| \((3\sqrt{5}-2\sqrt{3})^2\) | \(=\) | \(45-12\sqrt{15}+12\) |
| \(\) | \(=\) | \(\mathbf{57-12\sqrt{15}}\) |
(ii) This is \((a+b)(a-b) = a^2-b^2\): the middle terms cancel.
| \((4+\sqrt{7})(4-\sqrt{7})\) | \(=\) | \(16-7 = \mathbf{9}\) |
The answer is rational. Pairs like \(4+\sqrt{7}\) and \(4-\sqrt{7}\) are called conjugates.
Square first, then multiply by the bracket once more.
| \((2+\sqrt{5})^2\) | \(=\) | \(4+4\sqrt{5}+5 = 9+4\sqrt{5}\) |
| \((2+\sqrt{5})^3\) | \(=\) | \((9+4\sqrt{5})(2+\sqrt{5})\) |
| \(\) | \(=\) | \(18+9\sqrt{5}+8\sqrt{5}+20\) |
| \(\) | \(=\) | \(18+20+17\sqrt{5}\) |
| \(\) | \(=\) | \(\mathbf{38+17\sqrt{5}}\) |
So \(a = 38\) and \(b = 17\). \((2+\sqrt{5})^3\) is not \(8+5\sqrt{5}\): a power of a sum must be expanded.
(i) Expand \((2\sqrt{3}-3)^2\).
(ii) Hence simplify \(\sqrt{21-12\sqrt{3}}\).
(i)
| \((2\sqrt{3}-3)^2\) | \(=\) | \(12-12\sqrt{3}+9\) |
| \(\) | \(=\) | \(21-12\sqrt{3}\) |
(ii) So \(21-12\sqrt{3}\) is the square of \(2\sqrt{3}-3\), and also of \(3-2\sqrt{3}\). A square root is never negative, so choose the positive one. \(2\sqrt{3} = \sqrt{12} \gt 3\), so \(2\sqrt{3}-3 \gt 0\).
| \(\sqrt{21-12\sqrt{3}}\) | \(=\) | \(\mathbf{2\sqrt{3}-3}\) |
Check: \(2\sqrt{3}-3\approx 0.46\) and \(0.46^2\approx 0.21\approx 21-12\sqrt{3}\).
Common pitfalls
Frequently asked questions
How do you expand brackets with surds?
Multiply each term in the first bracket by each term in the second, the same as with pronumerals. Then simplify: a surd squared is a whole number, simplify each surd, and collect whole numbers and like surds.
What is the conjugate of a surd expression?
The conjugate of \(a+\sqrt{b}\) is \(a-\sqrt{b}\). Multiplying conjugates gives \(a^2-b\), which has no surd, so it is used to remove surds from denominators.
How do you square a binomial with surds?
Use \((a+b)^2 = a^2+2ab+b^2\). For \((\sqrt{5}+\sqrt{3})^2\) this gives \(5+2\sqrt{15}+3 = 8+2\sqrt{15}\).
What does the form a plus b root n mean?
It means a whole number part \(a\) plus a multiple \(b\) of a single surd \(\sqrt{n}\), such as \(19+18\sqrt{3}\). Collect all the rational terms into \(a\) and all the surd terms into \(b\).
How do you find unknowns by equating parts?
Expand the expression with the unknowns into a rational part plus a surd part. Set the rational parts equal and the surd parts equal, then solve the two equations.