Irrational numbers

The direct answer is option B, −10. Simplify the expression from inside the brackets. First, √40 = √(4 × 10) = 2√10. Hence 3√10 − 2√40 = 3√10 − 2(2√10) = 3√10 − 4√10 = −√10. Now multiply by the outside factor: √10(−√10) = −(√10)² = −10. Option A, 10, misses the negative sign and is therefore wrong. Option B, −10, is correct because the bracket becomes negative and the product of √10 with √10 is 10. Option C, 30 − 4√10, is wrong because it comes from incomplete or incorrect expansion and is not the simplified value. Option D, −2√40, is wrong because it leaves the expression unsimplified and also does not include the contribution of the first term correctly. The key rule is to simplify radicals before multiplying: take out perfect-square factors such as 4 from under √40.

Multiplying numerator and denominator by (\sqrt{19}) gives (\frac{6\sqrt{19}}{19}). Rationalisation removes the radical from the denominator.

(\sqrt{12}=2\sqrt{3}) and (\sqrt{108}=6\sqrt{3}), so the sum is (8\sqrt{3}). Find simplified forms before comparing.

It has no fixed repeating block, so it is irrational. Identifying non-terminating non-repeating decimals is important.

The direct answer is option A, (5 − √6)/19. To rationalise a denominator means to remove the square root from the denominator. The conjugate of 5 + √6 is 5 − √6, so multiply numerator and denominator by it: 1/(5 + √6) × (5 − √6)/(5 − √6). The denominator becomes (5 + √6)(5 − √6) = 25 − 6 = 19. Thus the result is (5 − √6)/19. Option A is correct because its denominator is rational and the fraction is equivalent to the original. Option B, (5 + √6)/19, is wrong because the conjugate sign was not used correctly; direct multiplication does not produce that numerator. Option C, 5 − √6, is wrong because it omits the denominator 19. Option D, 1/(5 − √6), is only the reciprocal-looking conjugate expression; it is not equal to the original fraction and does not rationalise its denominator. The exam cue is: for a denominator a + √b, multiply by a − √b, and use the difference of squares.

The number of zeros between successive 1s increases as 1, 2, 3, 4... Hence, no fixed block repeats and the number is irrational. Exam tip: a non-terminating, non-repeating decimal is irrational.

(\sqrt{48}=4\sqrt{3}) and (\sqrt{27}=3\sqrt{3}), so the sum is (7\sqrt{3}). Simplify before adding like radicals.

The numbers of zeros between successive 1s are 1, 2, 3, 4, …, so no fixed digit block can repeat forever. A non-terminating, non-repeating decimal is irrational. In exams, check repetition, not merely whether the decimal is infinite.

The governing concept is rationalisation of a denominator containing a surd. To remove √6 + 1 from the denominator, multiply the numerator and denominator by its conjugate, √6 − 1. Thus 1/(√6 + 1) × (√6 − 1)/(√6 − 1) = (√6 − 1)/[(√6)^2 − 1^2]. Applying the difference-of-squares identity gives the denominator 6 − 1 = 5, so the rationalised form is (√6 − 1)/5. Therefore option C is correct. Option A uses the original expression instead of the conjugate, so it does not produce the required rational denominator. Option B leaves the denominator irrational, and option D is not the equivalent result of multiplying by the appropriate conjugate. The value is unchanged; only its representation has been altered.

Since \(\sqrt{36}=6=\frac{6}{1}\), it is rational, so Riya’s statement is false. \(2\), \(5\), and \(11\) are not perfect squares, so their square roots are irrational. Exam tip: check for perfect squares first.

Both \(\sqrt{3}\) and \(-\sqrt{3}\) are irrational, but their sum is \(0\), a rational number. Hence the word “always” makes the claim false. Exam tip: one counterexample is enough to disprove a universal statement.

(\sqrt{98}=7\sqrt{2}) and (\sqrt{50}=5\sqrt{2}), so the difference is (2\sqrt{2}). First take out perfect-square factors.

(\sqrt{28}=2\sqrt{7}) and (\sqrt{63}=3\sqrt{7}), so the total is (6\sqrt{7}). Convert all terms to like radicals.

This decimal has no fixed repeating block, so it is irrational. Identify non-terminating non-repeating decimals.

Using the quotient property of square roots, \(\frac{\sqrt{108}}{\sqrt{3}}=\sqrt{\frac{108}{3}}=\sqrt{36}=6\). Therefore, option A is correct. Option D represents only \(\sqrt{108}=6\sqrt{3}\) and does not complete the division by \(\sqrt{3}\). Exam tip: For positive radicands, use \(\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\) to simplify such expressions.

(\sqrt{162}=9\sqrt{2}) and (\sqrt{72}=6\sqrt{2}), so the sum is (15\sqrt{2}). Add coefficients of like radicals.

Since (21<25<30), (\sqrt{25}) lies between them. Compare square roots using the numbers inside.

The zero blocks between successive 1s have lengths 1, 2, 3, 4, …, so no fixed block repeats. It is a non-terminating, non-recurring decimal and hence irrational. Exam tip: check repetition, not just the digits used.

\(0.\overline{27}=27/99=3/11\), so it is rational despite being non-terminating. Only non-terminating, non-repeating decimals are irrational. In exams, convert a repeating decimal into a fraction.

Multiplying numerator and denominator by (\sqrt{3}) gives (\frac{7\sqrt{3}}{3}). Rationalisation removes the radical from the denominator.

The governing concept is simplification of surds before applying an exponent. First, √12 = √(4 × 3) = 2√3. Therefore √12 + √3 = 2√3 + √3 = 3√3. Squaring gives (3√3)^2 = 3^2(√3)^2 = 9 × 3 = 27, so option B is correct. The result can also be checked by using (a + b)^2 = a^2 + 2ab + b^2, but simplifying the radicals first is more efficient and avoids unnecessary expansion. Option A may result from adding the radicands, option D may result from forgetting that the radical is also squared, and option C is only an unsimplified or incorrect partial expression, not the value of the complete square. Thus 27 is the unique correct value.

The zeros between 1s increase as 1, 2, 3, 4, …, so no fixed block repeats. A non-terminating, non-repeating decimal is irrational. Exam tip: check repetition, not merely which digits occur.

(\sqrt{147}=7\sqrt{3}), (\sqrt{75}=5\sqrt{3}) and (\sqrt{27}=3\sqrt{3}), so the answer is (9\sqrt{3}). Add and subtract coefficients of like radicals.

\(\frac{1}{3}=0.333\ldots\) has an infinite decimal expansion, but the digit 3 repeats, so it is rational. \(\sqrt{2}\) is non-terminating and non-repeating. Exam tip: recurring decimals are rational.

The square root of a positive integer is irrational when it is not a perfect square. Check for a perfect square first.