Radical and Surd

Unit 9 · Compulsory Mathematics · 11 min read

Radical and Surd

What the syllabus asks for 2
  • Radical and Surd: concepts, formulas and worked problems
  • Exercises of the SEE type

The diagonal of a square of side 1 cm is \sqrt{2} cm. No fraction and no terminating or repeating decimal equals \sqrt{2} exactly, so we leave it as a radical and calculate with it as a symbol. A surd is such an irrational root. This chapter shows how to classify roots, simplify them, add, subtract, multiply and divide them, and — the main skill for the exam — rationalise a denominator using a conjugate. These steps lead directly to the surds you meet later in trigonometry, coordinate geometry and quadratic equations.

What the SEE asks from this chapter

Usual questions: (i) simplify an expression with surds (3 marks); (ii) rationalise the denominator of a fraction such as \frac{3}{\sqrt{7}-\sqrt{5}} and simplify; (iii) if x=2+\sqrt{3}, find x+\frac{1}{x}, x^2+\frac{1}{x^2} or x^3+\frac{1}{x^3} (4 marks); (iv) find a and b when \frac{3+\sqrt{2}}{3-\sqrt{2}}=a+b\sqrt{2}; (v) solve a simple equation containing a square root. Always check answers to radical equations in the original equation.

9.1 Radical and Surd: concepts, formulas and worked problems

Radicals and surds

Radical, index and radicand

An expression \sqrt[n]{a} is called a radical. Here n is the index (or order) of the radical, a is the radicand, and \sqrt{\;} is the radical sign. The radical \sqrt[n]{a} means the number whose n-th power is a, so \left(\sqrt[n]{a}\right)^n=a. When n=2 it is written \sqrt{a} (square root); \sqrt[3]{a} is the cube root.

Surd

A surd is a radical \sqrt[n]{a} where a is a positive rational number and the value of the root is irrational (it cannot be written as a fraction).

Radical Value Surd?
\sqrt{16} 4 No — rational
\sqrt{18} 3\sqrt{2} Yes — irrational
\sqrt[3]{27} 3 No — rational
\sqrt[3]{5} 1.7099\ldots Yes
\sqrt{\frac{4}{9}} \frac{2}{3} No — rational
\sqrt{\pi} irrational but radicand \pi is not rational No — not a surd

Every surd is a radical, but not every radical is a surd.

\sqrt{2} is irrational

We assume that \sqrt{2} is rational and reach a contradiction. So \sqrt{2} cannot be rational.

Setting up. Suppose \sqrt{2}=\dfrac pq, where p and q are whole numbers with no common factor (the fraction is in its lowest terms).

Step 1 — square both sides.

2=\frac{p^2}{q^2} \;\Rightarrow\; p^2=2q^2

Step 2 — p is even. p^2 equals 2q^2, which is even. A square is even only if the number itself is even, so p=2k for some whole number k.

Step 3 — substitute p=2k.

(2k)^2=2q^2 \;\Rightarrow\; 4k^2=2q^2 \;\Rightarrow\; q^2=2k^2

Step 4 — q is even too. By the same reasoning q is even.

Result. Both p and q are even, so they share the factor 2. This contradicts "no common factor". Hence \sqrt{2} is not rational.

What it means. \sqrt{2} is a non-terminating, non-repeating decimal 1.41421356\ldots. It can be located on the number line, but only by geometry (figure below), not as a fraction.

figure
Figure 1. √2 is the diagonal of a unit square. Swinging the diagonal down to the number line with a compass marks the exact point √2 ≈ 1.414.

Laws of surds

Let a,b be positive rational numbers and m,n positive integers.

Laws of radicals

\sqrt[n]{a}\times\sqrt[n]{b}=\sqrt[n]{ab} \qquad \frac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\frac ab} \qquad \left(\sqrt[n]{a}\right)^n=a

\sqrt[n]{a^m}=a^{m/n} \qquad \sqrt[m]{\sqrt[n]{a}}=\sqrt[mn]{a}

Note: \sqrt{a+b}\ne\sqrt{a}+\sqrt{b} and \sqrt{a-b}\ne\sqrt{a}-\sqrt{b}. For example \sqrt{9+16}=5, but \sqrt{9}+\sqrt{16}=7.

Types of surds

Term Meaning Example
Pure (entire) surd no rational factor outside the radical \sqrt{18}, \sqrt[3]{24}
Mixed surd a rational number times a surd 3\sqrt{2}, 2\sqrt[3]{3}
Simple (monomial) surd a single term 5\sqrt{3}
Binomial surd two terms, at least one a surd 2+\sqrt{3}, \sqrt{5}-\sqrt{2}
Like (similar) surds same radicand and same order 2\sqrt{3} and 7\sqrt{3}
Unlike surds different radicand or order \sqrt{2} and \sqrt{3}
Simplifying radicals

Problem. Express as a mixed surd in simplest form: (a) \sqrt{72}, (b) \sqrt[3]{54}.

Solution. Take out the largest perfect square (or cube) factor.

(a) \sqrt{72}=\sqrt{36\times 2}=\sqrt{36}\times\sqrt{2}=6\sqrt{2}.

(b) \sqrt[3]{54}=\sqrt[3]{27\times 2}=\sqrt[3]{27}\times\sqrt[3]{2}=3\sqrt[3]{2}.

Mixed surd to pure surd

Problem. Express 3\sqrt{2} and 2\sqrt[3]{3} as entire (pure) surds.

Solution. Put the number inside the radical raised to the index.

3\sqrt{2}=\sqrt{3^2}\times\sqrt{2}=\sqrt{9\times 2}=\sqrt{18}.

2\sqrt[3]{3}=\sqrt[3]{2^3\times 3}=\sqrt[3]{24}.

Addition and subtraction

Only like surds can be added or subtracted: add or subtract the numbers in front, keep the surd. Unlike surds must first be simplified; if they are still unlike, leave the expression as it is.

Adding and subtracting surds

Problem. Simplify (a) \sqrt{50}+\sqrt{18}-\sqrt{8}, (b) 3\sqrt{12}-\sqrt{27}.

Solution. (a) \sqrt{50}=5\sqrt{2}, \sqrt{18}=3\sqrt{2}, \sqrt{8}=2\sqrt{2}.

\sqrt{50}+\sqrt{18}-\sqrt{8}=5\sqrt{2}+3\sqrt{2}-2\sqrt{2}=6\sqrt{2}

(b) 3\sqrt{12}=3\times 2\sqrt{3}=6\sqrt{3} and \sqrt{27}=3\sqrt{3}.

3\sqrt{12}-\sqrt{27}=6\sqrt{3}-3\sqrt{3}=3\sqrt{3}

Common mistakes
  • \sqrt{2}+\sqrt{3}=\sqrt{5} is wrong. Unlike surds cannot be combined: \sqrt{2}+\sqrt{3}\approx 3.146 but \sqrt{5}\approx 2.236.
  • \sqrt{a+b}=\sqrt{a}+\sqrt{b} is wrong. Square roots do not split over addition.
  • Stopping at \sqrt{72} when the answer should be 6\sqrt{2}. Always simplify to the mixed form.
  • Forgetting to simplify before adding: \sqrt{50} and \sqrt{18} look unlike but are both multiples of \sqrt{2}.

Multiplication and division

For like orders, multiply the numbers outside together and the numbers inside together.

Multiplication

Problem. Simplify (a) \sqrt{6}\times\sqrt{15}, (b) (2\sqrt{3}+\sqrt{5})(2\sqrt{3}-\sqrt{5}), (c) (\sqrt{5}+\sqrt{3})^2.

Solution. (a) \sqrt{6}\times\sqrt{15}=\sqrt{90}=\sqrt{9\times 10}=3\sqrt{10}.

(b) Use (a+b)(a-b)=a^2-b^2 with a=2\sqrt{3}, b=\sqrt{5}:

(2\sqrt{3})^2-(\sqrt{5})^2=4\times 3-5=12-5=7

(c) Use (a+b)^2=a^2+2ab+b^2:

(\sqrt{5})^2+2\sqrt{5}\sqrt{3}+(\sqrt{3})^2=5+2\sqrt{15}+3=8+2\sqrt{15}

The identity (a+b)(a-b)=a^2-b^2

When a binomial surd is multiplied by its conjugate (the same two terms with the sign between them reversed), the surds disappear and the result is rational: (3+\sqrt{2})(3-\sqrt{2})=9-2=7.

Comparing surds of different order

To compare \sqrt[m]{a} and \sqrt[n]{b}, first write both with a common index (the LCM of m and n), then compare the radicands.

figure
Figure 2. Position on the number line: √3 ≈ 1.732 lies to the right of the cube root of 5 ≈ 1.710, although it is a smaller-looking radicand.

Which is greater?

Problem. Which is greater, \sqrt{3} or \sqrt[3]{5}?

Solution. The orders are 2 and 3; their LCM is 6.

\sqrt{3}=3^{1/2}=3^{3/6}=\sqrt[6]{3^3}=\sqrt[6]{27}

\sqrt[3]{5}=5^{1/3}=5^{2/6}=\sqrt[6]{5^2}=\sqrt[6]{25}

Since 27>25, \sqrt[6]{27}>\sqrt[6]{25}. Therefore \sqrt{3}>\sqrt[3]{5}.

(Check by decimals: \sqrt{3}=1.732 and \sqrt[3]{5}=1.710 ✓.)

Rationalisation

A denominator containing a surd is awkward to work with and to evaluate. Rationalising means multiplying the numerator and the denominator by a suitable surd so that the denominator becomes rational.

Conjugate surds

The conjugate of a+\sqrt{b} is a-\sqrt{b} (and the conjugate of \sqrt{a}+\sqrt{b} is \sqrt{a}-\sqrt{b}). The product of conjugates is always rational:

(a+\sqrt{b})(a-\sqrt{b})=a^2-b \qquad (\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})=a-b

How to rationalise
Denominator Multiply top and bottom by
\sqrt{a} \sqrt{a} (since \sqrt{a}\times\sqrt{a}=a)
a+\sqrt{b} its conjugate a-\sqrt{b}
\sqrt{a}+\sqrt{b} its conjugate \sqrt{a}-\sqrt{b}
\sqrt{a}-\sqrt{b} its conjugate \sqrt{a}+\sqrt{b}

Multiplying numerator and denominator by the same non-zero number does not change the value of the fraction.

Monomial denominator

Problem. Rationalise the denominator: (a) \dfrac{1}{\sqrt{3}}, (b) \dfrac{6}{2\sqrt{3}}.

Solution. (a) \dfrac{1}{\sqrt{3}}\times\dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}.

(b) \dfrac{6}{2\sqrt{3}}=\dfrac{3}{\sqrt{3}}=\dfrac{3\sqrt{3}}{3}=\sqrt{3}.

Binomial denominator

Problem. Rationalise: (a) \dfrac{1}{\sqrt{5}-2}, (b) \dfrac{3}{\sqrt{7}-\sqrt{5}}, (c) \dfrac{5}{3+\sqrt{7}}.

Solution. (a) Multiply by the conjugate \sqrt{5}+2:

\frac{1}{\sqrt{5}-2}\times\frac{\sqrt{5}+2}{\sqrt{5}+2}=\frac{\sqrt{5}+2}{5-4}=\sqrt{5}+2

(b) Multiply by \sqrt{7}+\sqrt{5}:

\frac{3(\sqrt{7}+\sqrt{5})}{(\sqrt{7})^2-(\sqrt{5})^2}=\frac{3(\sqrt{7}+\sqrt{5})}{7-5}=\frac{3(\sqrt{7}+\sqrt{5})}{2}

(c) Multiply by 3-\sqrt{7}:

\frac{5(3-\sqrt{7})}{9-7}=\frac{5(3-\sqrt{7})}{2}

A fraction with two binomials

Problem. Simplify \dfrac{\sqrt{3}+1}{\sqrt{3}-1}.

Solution. Multiply top and bottom by \sqrt{3}+1:

\frac{(\sqrt{3}+1)^2}{(\sqrt{3})^2-1^2}=\frac{3+2\sqrt{3}+1}{3-1}=\frac{4+2\sqrt{3}}{2}=2+\sqrt{3}

Sum of conjugate reciprocals

Problem. Show that \dfrac{1}{2+\sqrt{3}}+\dfrac{1}{2-\sqrt{3}}=4.

Solution. \dfrac{1}{2+\sqrt{3}}=\dfrac{2-\sqrt{3}}{4-3}=2-\sqrt{3} and \dfrac{1}{2-\sqrt{3}}=\dfrac{2+\sqrt{3}}{4-3}=2+\sqrt{3}.

Sum =(2-\sqrt{3})+(2+\sqrt{3})=4. ✓

Using decimal values

Problem. Given \sqrt{2}=1.414 and \sqrt{3}=1.732, find the value of \dfrac{1}{\sqrt{3}-\sqrt{2}} to 3 decimal places.

Solution. Rationalise first, because dividing by 0.318 would be awkward:

\frac{1}{\sqrt{3}-\sqrt{2}}=\frac{\sqrt{3}+\sqrt{2}}{3-2}=\sqrt{3}+\sqrt{2}=1.732+1.414=3.146

Values of expressions: x+\frac{1}{x} and its powers

If x=a+\sqrt{b} then \dfrac{1}{x} is its conjugate divided by a rational number, and often simply the conjugate. This makes the sum x+\frac{1}{x} a rational number, which can then be squared and cubed with the identities

Identities for x and \frac{1}{x}

x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2 \qquad x^2+\frac{1}{x^2}=\left(x-\frac{1}{x}\right)^2+2

x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)

Value of powers

Problem. If x=2+\sqrt{3}, find (a) \dfrac{1}{x}, (b) x+\dfrac{1}{x}, (c) x^2+\dfrac{1}{x^2}, (d) x^3+\dfrac{1}{x^3}.

Solution. (a) \dfrac{1}{x}=\dfrac{1}{2+\sqrt{3}}=\dfrac{2-\sqrt{3}}{4-3}=2-\sqrt{3}.

(b) x+\dfrac{1}{x}=(2+\sqrt{3})+(2-\sqrt{3})=4.

(c) x^2+\dfrac{1}{x^2}=\left(x+\dfrac{1}{x}\right)^2-2=4^2-2=14.

(d) x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)=4^3-3\times 4=64-12=52.

Check of (c): x-\dfrac{1}{x}=2\sqrt{3}, so x^2+\dfrac{1}{x^2}=(2\sqrt{3})^2+2=12+2=14 ✓.

Finding the rational and irrational parts

Problem. If \dfrac{3+\sqrt{2}}{3-\sqrt{2}}=a+b\sqrt{2} where a and b are rational, find a and b.

Solution. Multiply by the conjugate 3+\sqrt{2}:

\frac{(3+\sqrt{2})^2}{9-2}=\frac{9+6\sqrt{2}+2}{7}=\frac{11+6\sqrt{2}}{7}=\frac{11}{7}+\frac{6}{7}\sqrt{2}

Comparing rational and irrational parts: a=\dfrac{11}{7} and b=\dfrac{6}{7}.

Difference of two conjugate fractions

Problem. Show that \dfrac{\sqrt{5}+2}{\sqrt{5}-2}-\dfrac{\sqrt{5}-2}{\sqrt{5}+2}=8\sqrt{5}.

Solution. Rationalise each fraction. The denominator in both is (\sqrt{5})^2-2^2=5-4=1.

\frac{\sqrt{5}+2}{\sqrt{5}-2}=(\sqrt{5}+2)^2=5+4\sqrt{5}+4=9+4\sqrt{5}

\frac{\sqrt{5}-2}{\sqrt{5}+2}=(\sqrt{5}-2)^2=5-4\sqrt{5}+4=9-4\sqrt{5}

Subtracting: (9+4\sqrt{5})-(9-4\sqrt{5})=8\sqrt{5}. ✓

A square root of a binomial surd

Problem. Simplify \sqrt{3+2\sqrt{2}}.

Solution. Try to write 3+2\sqrt{2} as a perfect square (a+b)^2=a^2+b^2+2ab. We need a^2+b^2=3 and 2ab=2\sqrt{2}. Take a=1 and b=\sqrt{2}: then a^2+b^2=1+2=3 and 2ab=2\sqrt{2} ✓.

3+2\sqrt{2}=(1+\sqrt{2})^2 \;\Rightarrow\; \sqrt{3+2\sqrt{2}}=1+\sqrt{2}

Rectangle with surd sides

Problem. A rectangle has length (3+\sqrt{5}) cm and breadth (3-\sqrt{5}) cm. Find its area and perimeter.

Solution. Area =(3+\sqrt{5})(3-\sqrt{5})=9-5=4\text{ cm}^2.

Perimeter =2\left[(3+\sqrt{5})+(3-\sqrt{5})\right]=2\times 6=12 cm.

Both answers are rational, because the sides are conjugates.

Equations with square roots

To remove a square root, square both sides. This can introduce an extra (false) solution, so every answer must be checked in the original equation.

Simple radical equations

Problem. Solve (a) \sqrt{x+5}=7, (b) \sqrt{x+7}=x-5.

Solution. (a) Square both sides: x+5=49, so x=44. Check: \sqrt{44+5}=\sqrt{49}=7 ✓.

(b) Square both sides: x+7=(x-5)^2=x^2-10x+25.

x^2-11x+18=0 \;\Rightarrow\; (x-2)(x-9)=0 \;\Rightarrow\; x=2\ \text{or}\ x=9

Check x=9: LHS =\sqrt{16}=4, RHS =9-5=4 ✓.

Check x=2: LHS =\sqrt{9}=3, RHS =2-5=-3. These are not equal, so x=2 is rejected.

The solution is x=9.

Always check

A square root sign means the positive root. When you square an equation like \sqrt{x+7}=x-5, the false answer x=2 makes the right side negative. Substitute every answer back in; reject any that fail.

Surds in geometry

Surds appear whenever Pythagoras' theorem gives a non-perfect square.

Diagonal and height

Problem. (a) Find the diagonal of a square of side 5 cm. (b) Find the height of an equilateral triangle of side 6 cm.

Solution. (a) d=\sqrt{5^2+5^2}=\sqrt{50}=5\sqrt{2} cm \approx 7.07 cm.

(b) The height splits the base into 3 cm and 3 cm: h=\sqrt{6^2-3^2}=\sqrt{27}=3\sqrt{3} cm \approx 5.20 cm.

figure
Figure 3. The square-root spiral: each new right-angled triangle has one leg equal to 1 and the other leg equal to the previous hypotenuse, so the hypotenuses are √2, √3, √4, √5, √6, √7.

Quick surd values to remember

\sqrt{2}\approx 1.414, \sqrt{3}\approx 1.732, \sqrt{5}\approx 2.236, \sqrt{6}\approx 2.449, \sqrt{7}\approx 2.646. Rationalise before substituting a decimal value — it avoids dividing by a small number.

9.2 Exercises of the SEE type

Method

  1. Simplify each radical first: take out perfect-square factors.
  2. Add or subtract only like surds.
  3. Multiply conjugates with (a+b)(a-b)=a^2-b^2; expand squares with (a\pm b)^2.
  4. Rationalise by the conjugate of the denominator; simplify the numerator and cancel common factors.
  5. For x=a+\sqrt{b}: find \frac{1}{x}, then x+\frac{1}{x}, then use the identities for squares and cubes.
  6. For equations with a root: isolate the root, square, solve, and check.

Chapter summary

  • A surd \sqrt[n]{a} (a rational) is irrational; \sqrt{16}, \sqrt[3]{27} are not surds; \sqrt{2}, \sqrt[3]{5} are.
  • \sqrt{a}\times\sqrt{b}=\sqrt{ab}; \frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac ab}; (\sqrt{a})^2=a; but \sqrt{a+b}\ne\sqrt{a}+\sqrt{b}.
  • Add and subtract only like surds after simplifying: 5\sqrt{2}+3\sqrt{2}=8\sqrt{2}.
  • Conjugate of a+\sqrt{b} is a-\sqrt{b}; product =a^2-b (rational).
  • Rationalise \frac{1}{\sqrt{a}} by \sqrt{a}; rationalise \frac{1}{\sqrt{a}\pm\sqrt{b}} by \sqrt{a}\mp\sqrt{b}, giving denominator a-b.
  • If x=2+\sqrt{3} then \frac{1}{x}=2-\sqrt{3}, x+\frac{1}{x}=4, x^2+\frac{1}{x^2}=14, x^3+\frac{1}{x^3}=52.
  • x^2+\frac{1}{x^2}=(x+\frac{1}{x})^2-2 and x^3+\frac{1}{x^3}=(x+\frac{1}{x})^3-3(x+\frac{1}{x}).
  • Compare surds of different order by writing both with the LCM of the indices.
  • Radical equations: square both sides and check each answer.

Practice questions

Group A — Multiple choice (1 mark each)

  1. Which of the following is a surd? [1]a\sqrt{16}b\sqrt{18}c\sqrt[3]{8}d\sqrt{\frac{9}{4}}
  2. The rationalising factor of \sqrt{5} is [1]a\sqrt{5}b5c\sqrt{5}-1d\frac{1}{5}
  3. The conjugate of 3+\sqrt{2} is [1]a-3-\sqrt{2}b3-\sqrt{2}c\sqrt{2}-3d3+\sqrt{2}
  4. The simplest form of \sqrt{50} is [1]a25\sqrt{2}b10\sqrt{5}c5\sqrt{2}d2\sqrt{5}
  5. The value of (\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2}) is [1]a5b1c\sqrt{6}d1+2\sqrt{6}
Answers to Group A

1. (b) — \sqrt{18}=3\sqrt{2} is irrational; the others equal 4, 2, \frac{3}{2}.

2. (a) — \sqrt{5}\times\sqrt{5}=5.

3. (b) — change the sign between the terms.

4. (c) — \sqrt{25\times 2}=5\sqrt{2}.

5. (b) — 3-2=1.

Group B — Short answer (2 marks each)

  1. Simplify \sqrt{72}. [2]
  2. Simplify \sqrt{12}+\sqrt{27}. [2]
  3. Rationalise the denominator of \dfrac{6}{\sqrt{3}}. [2]
  4. Express 3\sqrt{2} as an entire surd. [2]
Answers to Group B

1. \sqrt{36\times 2}=6\sqrt{2}.

2. 2\sqrt{3}+3\sqrt{3}=5\sqrt{3}.

3. \frac{6}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}=\frac{6\sqrt{3}}{3}=2\sqrt{3}.

4. 3\sqrt{2}=\sqrt{9\times 2}=\sqrt{18}.

Group C — Short answer (3 marks each)

  1. Rationalise the denominator of \dfrac{3}{\sqrt{7}-\sqrt{5}}. [3]
  2. Simplify (\sqrt{5}+\sqrt{3})^2 and (\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3}). [3]
  3. If x=2+\sqrt{3}, find x+\dfrac{1}{x}. [3]
  4. Simplify \sqrt{50}+\sqrt{18}-\sqrt{8}. [3]
Answers to Group C

1. \frac{3(\sqrt{7}+\sqrt{5})}{7-5}=\frac{3(\sqrt{7}+\sqrt{5})}{2}.

2. (\sqrt{5}+\sqrt{3})^2=5+2\sqrt{15}+3=8+2\sqrt{15}. (\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})=5-3=2.

3. \frac{1}{x}=\frac{1}{2+\sqrt{3}}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}. x+\frac{1}{x}=(2+\sqrt{3})+(2-\sqrt{3})=4.

4. 5\sqrt{2}+3\sqrt{2}-2\sqrt{2}=6\sqrt{2}.

Group D — Long answer (4 marks each)

  1. If x=3+2\sqrt{2}, find x+\dfrac{1}{x}, x^2+\dfrac{1}{x^2} and x^3+\dfrac{1}{x^3}. [4]
  2. Simplify \dfrac{\sqrt{3}+1}{\sqrt{3}-1}+\dfrac{\sqrt{3}-1}{\sqrt{3}+1}. [4]
  3. If \dfrac{3+\sqrt{2}}{3-\sqrt{2}}=a+b\sqrt{2}, find the rational numbers a and b. [4]
  4. Solve \sqrt{x+7}=x-5. [4]
Answers to Group D

1. \frac{1}{x}=\frac{1}{3+2\sqrt{2}}=\frac{3-2\sqrt{2}}{9-8}=3-2\sqrt{2}. x+\frac{1}{x}=6. x^2+\frac{1}{x^2}=6^2-2=34. x^3+\frac{1}{x^3}=6^3-3\times 6=216-18=198.

2. \frac{(\sqrt{3}+1)^2+(\sqrt{3}-1)^2}{(\sqrt{3})^2-1}=\frac{(4+2\sqrt{3})+(4-2\sqrt{3})}{2}=\frac{8}{2}=4.

3. \frac{(3+\sqrt{2})^2}{9-2}=\frac{11+6\sqrt{2}}{7}, so a=\frac{11}{7} and b=\frac{6}{7}.

4. Square both sides: x+7=x^2-10x+25, so x^2-11x+18=0 and x=2 or 9. Check x=9: \sqrt{16}=4=9-5 ✓. Check x=2: \sqrt{9}=3\ne 2-5=-3 ✗. So x=9.

Practise more: Mathematics question bank