Real Numbers

( \sqrt{27}>4 ) because (27>16). So (4-\sqrt{27}) is negative and the absolute value is ( \sqrt{27}-4 ).

\(\sqrt{2}\) is irrational, but \(\sqrt{2}\times\sqrt{2}=2\), which is rational. Therefore, the product of two irrational numbers is not always irrational. In option C, \(\sqrt{6}\) is irrational, so it does not disprove the claim. Exam tip: To disprove an “always” statement, one valid counterexample is enough.

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.

\(x+r\) 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.

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.

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.

First find ( (3+\sqrt{5})^2=14+6\sqrt{5} ). Multiplying by (3+\sqrt{5}) gives (72+32\sqrt{5}).

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

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.

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.

The governing concept is the use of a counterexample to disprove a statement claimed to hold for all allowed numbers. In option A, the left side is √(4 + 9) = √13, whereas the right side is √4 + √9 = 2 + 3 = 5. Since √13 is not equal to 5, this example directly proves that square roots do not distribute over addition. Options B, C, and D are true only because one of the addends is zero: adding zero leaves the other number unchanged. These special cases do not establish the proposed rule for general positive numbers. Therefore option A is the valid counterexample and the unique correct answer.

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

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.

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

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

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.

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.

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.

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.

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.