Real Numbers

Since \(320=64\times5\) and \(64\) is a perfect square, \(\sqrt{320}=\sqrt{64\times5}=\sqrt{64}\times\sqrt{5}=8\sqrt{5}\). In option B, the coefficient has been incorrectly doubled, while option C results from an incorrect factorisation of 320. Exam tip: take the largest perfect-square factor outside the square root.

The governing concept is simplifying a square root by taking the largest perfect-square factor outside the radical. Factor 363 as 121 × 3, and 121 is 11². Therefore √363 = √(121 × 3) = √121 × √3 = 11√3. The remaining factor 3 has no perfect-square factor greater than 1, so 11√3 is the simplest form. Thus option B is correct. Option A comes from an incorrect factorisation of 363, and option C reverses the coefficient and radical structure. Option D is not equivalent because squaring 9√33 gives 81×33, not 363. The principal square root is positive, so the answer is 11√3 rather than its negative.

\(432=144\times3=12^2\times3\). Therefore, \(\sqrt{432}=\sqrt{12^2\times3}=12\sqrt{3}\), so option B is correct. Option A does not use the correct perfect-square factor, while squaring options C and D does not give 432. Exam tip: factor the number using the largest perfect-square factor before simplifying a surd.

\(\frac{7}{24}=0.29166\ldots\), where 6 repeats, so its decimal expansion is non-terminating recurring. Every non-terminating recurring decimal represents a rational number; only non-terminating non-recurring decimals are irrational. Option A wrongly ignores the difference between non-terminating and non-terminating recurring decimals. Exam tip: If a digit or group of digits repeats regularly, the number is rational.

In \(0.101001000100001\ldots\), the number of zeros between successive 1s keeps increasing, so no fixed block of digits repeats. Its decimal expansion is non-terminating and non-recurring; therefore, the number is irrational. Using only the digits 0 and 1 does not make a number rational. Exam tip: A rational number has either a terminating decimal expansion or a non-terminating recurring decimal expansion.

By distributive law, ( \sqrt{7}\times4+\sqrt{7}\times\sqrt{7}=4\sqrt{7}+7 ). Multiplying the same square root gives the radicand.

A non-terminating, non-recurring decimal expansion is a characteristic of an irrational number. For example, the decimal expansion of (\sqrt{2}) is 1.414213..., which neither terminates nor repeats in a fixed pattern. Options B and D are incorrect because non-terminating recurring decimals such as 0.333... are rational. Exam tip: Terminating or recurring decimals are rational, while non-terminating non-recurring decimals are irrational.

The governing concept is the algebraic identity (a − b)² = a² − 2ab + b². Here a = 7 and b = √3. Substituting these values gives (7 − √3)² = 7² − 2(7)(√3) + (√3)². Now 7² = 49 and (√3)² = 3, so the expression becomes 49 − 14√3 + 3 = 52 − 14√3. Hence option A is correct. Option B has an incorrect constant and misses the correct middle coefficient. Option C changes √3 to √7 without any mathematical reason. Option D obtains the correct constant 52 but incorrectly uses 7√3 instead of 14√3. The factor 2 in the middle term is essential and is the most common source of error in this expansion.

The two factors are conjugates. Using \((a+b)(a-b)=a^2-b^2\), we get \((6+\sqrt{23})(6-\sqrt{23})=6^2-(\sqrt{23})^2=36-23=13\). Option A results from incorrectly adding 23. Exam tip: When conjugate factors appear, apply the difference of squares directly.

\(\sqrt{8}=\sqrt{4\times2}=2\sqrt{2}\). Therefore, \(\sqrt{2}+\sqrt{8}=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\). Since \(\sqrt{2}\) is irrational, its product with a non-zero rational number such as 3 is also irrational. Hence the sum is irrational. Option C uses the incorrect rule \(\sqrt{a}+\sqrt{b}=\sqrt{a+b}\). Exam tip: simplify each surd first and then combine like surds.

To rationalise the denominator, multiply both numerator and denominator by \sqrt{11}: \frac{8}{\sqrt{11}}\times\frac{\sqrt{11}}{\sqrt{11}}=\frac{8\sqrt{11}}{11}. Hence, option A is correct. Option B is not equivalent to the original fraction, while option C omits the factor \sqrt{11} in the numerator. Exam tip: for a denominator containing a single square root, multiply by the same square root to obtain a perfect-square denominator.

Multiplying numerator and denominator by ( \sqrt{5} ) gives ( \frac{9\sqrt{5}}{20} ). The radical should be removed from the denominator.

The denominator is ( \sqrt{13}+2\sqrt{3} ) and its conjugate is ( \sqrt{13}-2\sqrt{3} ). The denominator becomes (13-12=1).

To rationalise the denominator, multiply the numerator and denominator by the conjugate \(\sqrt{17}+\sqrt{8}\). The denominator becomes \((\sqrt{17}-\sqrt{8})(\sqrt{17}+\sqrt{8})=17-8=9\), while the numerator becomes \(\sqrt{17}+\sqrt{8}\). Hence, the correct form is \(\frac{\sqrt{17}+\sqrt{8}}{9}\). Exam tip: use the identity \((a-b)(a+b)=a^2-b^2\).

( \sqrt{128}=8\sqrt{2} ), ( \sqrt{200}=10\sqrt{2} ), and ( \sqrt{72}=6\sqrt{2} ). So the result is (12\sqrt{2}).

Since \(54=9\times6\) and \(150=25\times6\), we have \(\sqrt{54}=3\sqrt{6}\) and \(\sqrt{150}=5\sqrt{6}\). Therefore, \(3\sqrt{54}+2\sqrt{150}=3(3\sqrt{6})+2(5\sqrt{6})=9\sqrt{6}+10\sqrt{6}=19\sqrt{6}\). Hence, option B is correct. Exam tip: Radicals with the same radicand can be added or subtracted by combining their coefficients.

In option C, the number of zeros between successive 1s keeps increasing. Its decimal expansion neither terminates nor repeats a fixed pattern, so it is irrational. Option B is non-terminating, but the digit 3 repeats, making it a recurring decimal and therefore rational. Exam tip: A non-terminating, non-repeating decimal is always irrational, while a terminating or recurring decimal is rational.

The decimal expansion of an irrational number is non-terminating, but no fixed block of digits repeats indefinitely; hence it is non-recurring. For example, \(\sqrt{2}=1.414213\ldots\) is non-terminating and non-recurring. In contrast, \(1/3=0.333\ldots\) is non-terminating and recurring, so it is rational. Exam tip: recurring decimals are rational, while non-terminating non-recurring decimals are irrational.

The governing concept is the product property of square roots: for non-negative real numbers, √a × √b = √(ab). Thus √45 × √80 = √(45 × 80) = √3600. Because 3600 = 60² and a principal square root is non-negative, √3600 = 60, making option A correct. The result can also be checked by simplifying separately: √45 = √(9 × 5) = 3√5 and √80 = √(16 × 5) = 4√5. Their product is (3√5)(4√5) = 12 × 5 = 60. Option B is only half of the correct result, option C does not simplify to 60, and option D uses an incorrect radicand. Therefore A is the only equivalent value.

( \frac{\sqrt{242}}{\sqrt{2}}=\sqrt{121}=11 ). In division of square roots, divide the radicands.