Real Numbers

Correct answer: B. The statement is incorrect; the correct result is \(3\sqrt{2}\), which is irrational.

Explanation: 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, \(\sqrt{a}+\sqrt{b}\neq\sqrt{a+b}\).

Correct answer: A. 5

Explanation: \(\sqrt{25}=5\) 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.

Correct answer: B. \(9\)

Explanation: 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
\(x\).

Correct answer: A. 0

Explanation: 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 \(|x|\).

Correct answer: D. (5-2\sqrt{6})

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

Correct answer: A. \(8\sqrt{5}\)

Explanation: 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.

Correct answer: B. \(12\sqrt{3}\)

Explanation: \(432=144\times3=12^2\times3\). 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.

Correct answer: B. A number with a non-terminating recurring decimal expansion is rational.

Explanation: \(\frac{7}{24}=0.29166\ldots\), 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.

Correct answer: C. It is irrational because its decimal expansion is neither terminating nor recurring.

Explanation: 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.

Correct answer: B. (4\sqrt{7}+7)

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

Correct answer: C. A non-terminating non-recurring decimal expansion represents an irrational number.

Explanation: 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.

Correct answer: A. 52 − 14√3

Explanation: 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.

Correct answer: B. 13

Explanation: 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.

Correct answer: A. Since \(\sqrt{8}=2\sqrt{2}\), the sum is \(3\sqrt{2}\), which is irrational.

Explanation: \(\sqrt{8}=\sqrt{4\times2}=2\sqrt{2}\). 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.

Correct answer: A. \frac{8\sqrt{11}}{11}

Explanation: 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.

Correct answer: A. ( \frac{9\sqrt{5}}{20} )

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

Correct answer: A. ( \sqrt{13}-2\sqrt{3} )

Explanation: 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.

Correct answer: A. \(\frac{\sqrt{17}+\sqrt{8}}{9}\)

Explanation: 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 \((a-b)(a+b)=a^2-b^2\).

Correct answer: A. (12\sqrt{2})

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

Correct answer: B. \(19\sqrt{6}\)

Explanation: 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.

Correct answer: C. 0.101001000100001...

Explanation: 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.

Correct answer: B. The decimal expansion of irrational numbers is non-terminating and non-recurring.

Explanation: 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.

Correct answer: B. (11)

Explanation: 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.

Correct answer: C. \(\frac{11}{12}\)

Explanation: 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.

Correct answer: B. \(\sqrt{5}+(-\sqrt{5})\)

Explanation: 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.