Real Numbers

Since \(\sqrt{8}=\sqrt{4\times2}=2\sqrt{2}\), we get \(\sqrt{2}+\sqrt{8}=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\), not \(\sqrt{10}\). Because \(\sqrt{2}\) is irrational, its product with the non-zero rational number 3 is also irrational. Exam tip: simplify radicals first and combine only like radical terms; in general, \(\sqrt{a}+\sqrt{b}\neq\sqrt{a+b}\).

\(\sqrt{25}=5\) and \(\sqrt{100}=10\). Therefore, \(\left|\sqrt{25}-\sqrt{100}\right|=|5-10|=|-5|=5\). Option B results from adding the two values instead of subtracting them, while option C ignores that an absolute value cannot be negative. Exam tip: the absolute value of any real number is always non-negative.

For real numbers,
\(\sqrt{x^2}=|x|\). Therefore,
\(\sqrt{(-9)^2}=|-9|=9\). The principal square root is always non-negative, so
\(-9\) is not the answer. Exam tip: simplify
\(\sqrt{x^2}\) as
\(|x|\), not automatically as
\(x\).

For every real number \(x\), \(\sqrt{x^2}=|x|\), not always \(x\). Thus, for \(x=-8\), \(\sqrt{x^2}+x=|-8|+(-8)=8-8=0\). Option B results from the common mistake of taking \(\sqrt{x^2}=x\) directly. Exam tip: In such questions, first rewrite \(\sqrt{x^2}\) as the absolute value \(|x|\).

Multiplying by the conjugate (5-2\sqrt{6}) makes the denominator (25-24=1). So the rationalised form is (5-2\sqrt{6}).

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.

\(432=144\times3=12^2\times3\). 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 direct answer is option A: \(\sqrt{13}-2\sqrt{3}\). First rewrite \(\sqrt{12}=2\sqrt{3}\), so the fraction is \(1/(\sqrt{13}+2\sqrt{3})\). Multiply numerator and denominator by the conjugate \(\sqrt{13}-2\sqrt{3}\). The denominator becomes \((\sqrt{13})^2-(2\sqrt{3})^2=13-12=1\), so the result is \((\sqrt{13}-2\sqrt{3})/1\). Option A states this result. Option B uses the same-sign expression and does not rationalise the denominator. Option C divides by 25, although the denominator is 1. Option D merely leaves a radical denominator. The useful rule is to multiply by the conjugate and use difference of squares.

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 direct answer is B: \(11\). Both square roots have nonnegative values, and \(\sqrt{a}/\sqrt{b}=\sqrt{a/b}\) when the denominator is valid. Therefore \(\sqrt{242}/\sqrt2=\sqrt{242/2}=\sqrt{121}=11\). Equivalently, \(242=2\times121\), so \(\sqrt{242}=\sqrt2\times11\), and division by \(\sqrt2\) leaves 11. Option B is correct. Option A, 121, is the radicand after division, not its square root. Option C, \(\sqrt{240}\), comes from an incorrect subtraction-like operation and is not equal to the expression. Option D, 22, doubles the correct result and has no valid rule behind it. Remember that \(\sqrt{121}=11\), not 121. A useful check is numerical: \(\sqrt{242}\) is about 15.56 and \(\sqrt2\) about 1.41; their quotient is about 11. Memory cue: simplify inside the radical, then take the root.

For \(\frac{11}{12}\), the denominator is \(12=2^2\times3\). Since it contains the prime factor 3, its decimal expansion is non-terminating recurring: \(0.91\overline{6}\). In contrast, the denominators 8, 20 and 125 have only 2 and/or 5 as prime factors, so their decimal expansions terminate. Exam tip: First reduce a fraction to its simplest form, then check the prime factors of its denominator.

Both \(\sqrt{5}\) and \(-\sqrt{5}\) are irrational, but \(\sqrt{5}+(-\sqrt{5})=0\). Since 0 is rational, this is a counterexample to the claim that the sum of two irrational numbers is always irrational. Option A does give an irrational sum, but it does not disprove an “always” statement. Exam tip: one counterexample is enough to disprove a universal statement.