Real Numbers

Correct answer: A. Both are equal

Explanation: Since \(32=16\times2\) and \(18=9\times2\), \(\sqrt{32}=4\sqrt{2}\) and \(\sqrt{18}=3\sqrt{2}\). Therefore, \(\sqrt{32}+\sqrt{18}=4\sqrt{2}+3\sqrt{2}=7\sqrt{2}\). Hence, both expressions are equal. Option D is incorrect because \(\sqrt{2}\), and therefore \(7\sqrt{2}\), is irrational. Exam tip: Before comparing surds, simplify each radical by taking out perfect-square factors.

Correct answer: A. \(x+r\) is irrational

Explanation: \(x+r\) must always be irrational. If \(x+r\) were rational, then subtracting the rational number \(r\) would make \(x=(x+r)-r\) rational, contradicting the fact that \(x\) is irrational. Option C is not always true because if \(r=0\), then \(xr=0\), which is rational. Exam tip: Adding or subtracting a rational number from an irrational number always gives an irrational number.

Correct answer: B. The product of a non-zero rational number and an irrational number is always irrational.

Explanation: Let \(r\) be a non-zero rational number and \(x\) be irrational. If \(rx\) were rational, then \(x=\frac{rx}{r}\) would also be rational, which is a contradiction. Hence, their product is always irrational. Options A and C are not always true: \(\sqrt{2}+(-\sqrt{2})=0\) and \(\sqrt{2}\times\sqrt{2}=2\), both rational. Exam tip: For a statement containing “always”, test it with a suitable counterexample.

Correct answer: A. 2√13/9

Explanation: Use the common-denominator rule and the difference-of-squares identity. The product of the denominators is (√13+2)(√13−2)=(√13)²−2²=13−4=9. When the fractions are added, the numerator is the sum of the opposite denominator parts: (√13−2)+(√13+2)=2√13. Hence the complete expression is (2√13)/9, which is option A. The irrational terms do not cancel here; they add because both terms have the same sign after the denominators are combined. Option B omits the denominator 9. Option C changes the factor incorrectly and has no valid algebraic basis. Option D treats √13 as if it could be cancelled without accounting for the conjugate product. Both original denominators are nonzero, so the calculation is defined.

Correct answer: A. (72+32\sqrt{5})

Explanation: The direct answer is option A: \(72+32\sqrt{5}\). Begin with \((3+\sqrt{5})^3=(3+\sqrt{5})^2(3+\sqrt{5})\). The square is \(3^2+2(3)(\sqrt{5})+(\sqrt{5})^2=9+6\sqrt{5}+5=14+6\sqrt{5}\). Now multiply: \((14+6\sqrt{5})(3+\sqrt{5})=42+14\sqrt{5}+18\sqrt{5}+30=72+32\sqrt{5}\). Option A is correct. Option B, \(42+18\sqrt{5}\), contains only part of the multiplication and misses terms. Option C, \(72+27\sqrt{5}\), has the wrong coefficient of \(\sqrt{5}\); the two radical terms add to \(32\sqrt{5}\). Option D, \(27+15\sqrt{5}\), is not the result of cubing and misses much of the expansion. A useful check is to keep rational terms and radical terms separate until the final addition.

Correct answer: A. (100-51\sqrt{3})

Explanation: First find ( (4-\sqrt{3})^2=19-8\sqrt{3} ). Multiplying by (4-\sqrt{3}) gives (100-51\sqrt{3}).

Correct answer: A. \(324<n<361\)

Explanation: We are given \(18<\sqrt{n}<19\). Since 18 and 19 are positive, squaring all parts preserves the inequality: \(18^2<n<19^2\). Hence, \(324<n<361\). In \(289<n<324\), \(\sqrt{n}\) would lie between 17 and 18, not between 18 and 19. Exam tip: Before squaring an inequality involving square roots, check that the bounds are positive.

Correct answer: A. 441<m<484

Explanation: Since \(\sqrt{m}\) is positive, squaring all parts preserves the inequality: \(21^2<m<22^2\). Thus \(441<m<484\) is correct. Closest distractors: option B (\(21<m<22\)) gives the range of \(\sqrt{m}\), not of \(m\); option C (\(400<m<441\)) equals \(20^2\)–\(21^2\) and option D (\(484<m<529\)) equals \(22^2\)–\(23^2\), so both are outside the required range. Exam tip: when squaring an inequality ensure the expressions are nonnegative — then you can square termwise without flipping the inequality sign; square the exact endpoints to get the correct interval.

Correct answer: A. (ab)

Explanation: Multiplying the square root of a non-negative number by itself gives the same number. Therefore the value is (ab).

Correct answer: A. \(\sqrt{36}+\sqrt{49}>\sqrt{85}\)

Explanation: Since \(\sqrt{36}=6\) and \(\sqrt{49}=7\), we get \(\sqrt{36}+\sqrt{49}=13\). Also, \(85<169=13^2\), so \(\sqrt{85}<13\). Therefore, \(\sqrt{36}+\sqrt{49}>\sqrt{85}\) is correct. Option B is incorrect because the two expressions are not equal. Exam tip: To compare positive square roots, compare the corresponding squared values.

Correct answer: A. ( \sqrt{3}\times\sqrt{27} )

Explanation: ( \sqrt{3}\times\sqrt{27}=\sqrt{81}=9 ), which is rational. The other products do not become square roots of perfect squares.

Correct answer: B. Irrational

Explanation: The decimal neither terminates nor repeats so it is irrational. Remember the difference between repeating and non repeating decimals.

Correct answer: C. \(5\sqrt{3}\)

Explanation: Since \(27=9\times3\), \(\sqrt{27}=\sqrt{9\times3}=3\sqrt{3}\). Therefore, \(2\sqrt{3}+\sqrt{27}=2\sqrt{3}+3\sqrt{3}=5\sqrt{3}\). The option \(4\sqrt{3}\) can result from simplifying \(\sqrt{27}\) incorrectly. In exams, first factor out the largest perfect square from inside a surd.

Correct answer: A. The statement is true; every irrational number has a non-terminating, non-repeating decimal expansion.

Explanation: The statement is true. A rational number has either a terminating decimal expansion or a non-terminating recurring decimal expansion. Therefore, a decimal that is non-terminating and non-repeating represents an irrational number, for example \(\sqrt{2}=1.414213\ldots\). Option D is wrong because \(1/3=0.333\ldots\) is non-terminating but rational, since it repeats. Exam tip: For decimal-expansion questions, always check whether the digits repeat.

Correct answer: C. 5

Explanation: Given \(x=\sqrt{5}\), we get \(x^2=(\sqrt{5})^2=5\). Squaring the principal square root of a positive number gives the number itself. \(25\) would result from squaring \(5\), not \(\sqrt{5}\). Exam tip: apply \((\sqrt{a})^2=a\) directly.

Correct answer: A. \(5\sqrt{3}\)

Explanation: Since \(12=4\times3\) and \(27=9\times3\), \(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{27}=3\sqrt{3}\). Therefore, \(\sqrt{12}+\sqrt{27}=2\sqrt{3}+3\sqrt{3}=5\sqrt{3}\). \(3\sqrt{3}\) is only the simplified form of \(\sqrt{27}\), not of the sum. Exam tip: first take out perfect-square factors from each surd, then add like surds.

Correct answer: A. \(\frac{1}{3}=0.333\ldots\)

Explanation: The decimal expansion of \(\frac{1}{3}=0.333\ldots\) is non-terminating, but it repeats. Since it can be written as a ratio of integers, \(\frac{1}{3}\), it is rational. In contrast, \(\sqrt{2}\), \(\pi\), and \(\sqrt{5}\) have non-terminating, non-repeating decimal expansions and are irrational. Exam tip: A non-terminating decimal is irrational only when it is non-repeating.

Correct answer: B. \(3\sqrt{2}\)

Explanation: \(\sqrt{8}=\sqrt{4\times2}=2\sqrt{2}\). Therefore, \(a+b=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\), so option B is correct. \(2\sqrt{2}\) is only the simplified value of \(b\), not the sum. Exam tip: simplify surds first, then add like surd terms.

Correct answer: A. \(2\sqrt{5}\)

Explanation: Since \(45=9\times5\), \(\sqrt{45}=\sqrt{9\times5}=3\sqrt{5}\). Therefore, \(\sqrt{45}-\sqrt{5}=3\sqrt{5}-\sqrt{5}=2\sqrt{5}\). \(3\sqrt{5}\) is only the simplified form of \(\sqrt{45}\), not the result after subtraction. Exam tip: before adding or subtracting surds, first extract perfect-square factors from the radicands.

Correct answer: C. Every real number is rational

Explanation: Statement C is false because the set of real numbers includes both rational and irrational numbers. For example, \(\sqrt{2}\) is a real number, but it cannot be written in the form \(p/q\); hence, it is not rational. Statement D is correct because real numbers are the union of rational and irrational numbers. Exam tip: Examples such as \(\sqrt{2}\) and \(\pi\) help identify irrational real numbers.

Correct answer: B. 4

Explanation: Using the quotient rule for square roots, \(\frac{\sqrt{32}}{\sqrt{2}}=\sqrt{\frac{32}{2}}=\sqrt{16}=4\). Therefore, 4 is correct. The number 16 is only the value inside the square root; its square root must still be evaluated. Exam tip: use \(\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\) when \(b>0\).

Correct answer: A. The claim is incorrect; it is irrational because its decimal expansion is non-terminating and non-repeating.

Explanation: The blocks of 0s keep increasing in length, so no fixed block of digits repeats. A non-terminating, non-repeating decimal is irrational. Exam tip: check repetition, not merely the digits used.

Correct answer: C. \(2\sqrt{7}\)

Explanation: Both terms are like surds: \(\sqrt{7}+\sqrt{7}=1\sqrt{7}+1\sqrt{7}=(1+1)\sqrt{7}=2\sqrt{7}\). \(\sqrt{14}\) is incorrect because \(\sqrt{a}+\sqrt{b}\) cannot generally be written as \(\sqrt{a+b}\). Exam tip: add only the coefficients of like surds.

Correct answer: C. Integer

Explanation: (-4) is a negative integer, so it belongs to the set of integers. Natural numbers and whole numbers do not include negative numbers. Although (-4) is also a rational number, integers form the smaller set containing it. Exam tip: Remember the usual classification order: natural numbers, whole numbers, integers, rational numbers, and real numbers.

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

Explanation: Since \(48=16\times3\), \(\sqrt{48}=\sqrt{16\times3}=4\sqrt{3}\). Therefore, \(\sqrt{48}-2\sqrt{3}=4\sqrt{3}-2\sqrt{3}=2\sqrt{3}\), so option A is correct. \(4\sqrt{3}\) is only the simplified form of \(\sqrt{48}\); the subtraction of \(2\sqrt{3}\) must still be done. Exam tip: first identify the largest perfect-square factor inside a surd.