Irrational numbers and real numbers

(\sqrt{5}\times\sqrt{5}=5) is rational but (\sqrt{5}\times\sqrt{2}=\sqrt{10}) is irrational. So it depends on the case.

Multiply using distributive property: \(\sqrt{5}(2+\sqrt{5})=2\sqrt{5}+\sqrt{5}\cdot\sqrt{5}=2\sqrt{5}+5\). So the value is \(5+2\sqrt{5}\). Option B is close but has the wrong coefficient of \(\sqrt{5}\) (it shows \(1\sqrt{5}\) instead of \(2\sqrt{5}\)). Option C swaps the coefficients of the rational and irrational parts, and option D (1) is incorrect — that would arise only if numerator and denominator were identical or by an invalid cancellation. Exam tip: distribute the surd across each term and then combine like (rational and irrational) terms carefully.

Use the difference-of-squares identity \((a+b)(a-b)=a^2-b^2\). With \(a=\sqrt{7}\) and \(b=1\) we get \(7-1=6\), so A is correct. Option B (8) is a common mistake from adding \(7+1\) instead of subtracting. Options C and D are different expressions and do not equal 6. Exam tip: spot the pattern \((a+b)(a-b)\) and apply \(a^2-b^2\) immediately to avoid arithmetic mistakes.

Use the square formula \\((a+b)^2=a^2+2ab+b^2\\). With \(a=2\) and \(b=\sqrt{3}\): \(a^2=4,\;2ab=4\sqrt{3},\;b^2=3\). Adding gives \(4+4\sqrt{3}+3=7+4\sqrt{3}\), so option A is correct. Option C (\(7+2\sqrt{3}\)) results from omitting the factor 2 in the middle term; option B (\(7\)) omits the middle term entirely; option D (\(4+4\sqrt{3}\)) omits the \(b^2\) term. Exam tip: compute each term \(a^2,\;2ab,\;b^2\) separately and optionally verify by approximating numerically.

\(\sqrt{13}\) is correct because \(9<13<16\) implies \(3<\sqrt{13}<4\), and 13 is not a perfect square so \(\sqrt{13}\) is irrational. The other choices fail: \(22/7\) and \(7/2\) are ratios of integers (hence rational), and \(\sqrt{16}=4\) is not between 3 and 4. Exam tip: For \(\sqrt{n}\) check whether n lies between two consecutive perfect squares to locate it, and test rationality by seeing if the number can be written as a fraction of integers or is a square root of a non-square integer.

(\frac{\sqrt{3}}{3}) is irrational and lies between (0) and (1). Dividing by a non zero rational keeps irrationality.

The square root of a positive integer is rational only when the number is a perfect square. This is a direct exam rule.

Let \(x = 2.\overline{45}\). The repeating block has length 2, so multiply by 100: \(100x = 245.\overline{45}\). Subtracting gives \(99x = 243\), hence \(x = \frac{243}{99} = \frac{27}{11}\). This is a fraction, so the number is rational. Option B (irrational) is incorrect because irrational numbers do not have terminating or repeating decimal expansions; here the decimal repeats. Option C (non-real) is wrong because the number is real, and option D (whole number) is wrong because there is a nonzero fractional part. Exam tip: An overline means a repeating block — use multiplication by \(10^n\) (where n is block length) to convert to a fraction quickly.

\(\sqrt{162}=\sqrt{81\times2}=\sqrt{81}\cdot\sqrt{2}=9\sqrt{2}\). Therefore the correct simplified form is \(9\sqrt{2}\). Option B (\(6\sqrt{3}\)) is incorrect because \(6\sqrt{3}=\sqrt{36\times3}=\sqrt{108}\), which is not equal to \(\sqrt{162}\). Options C and D are also numerically different (\(18=\sqrt{324}\), and \(18\sqrt{2}\) is much larger). Exam tip: factor out the largest perfect square from under the root to simplify quickly.

Simplify first: \(\sqrt{27}=\sqrt{9\cdot3}=3\sqrt{3}\). Thus \(4\sqrt{3}-\sqrt{27}=4\sqrt{3}-3\sqrt{3}=\sqrt{3}\), so option A is correct. Option B (1) is wrong because \(\sqrt{3}\approx1.732\), not 1. Option C (\(3\sqrt{3}\)) is incorrect because it treats the second term as if it should be added or left unsimplified; the correct simplification gives subtraction of like terms. Exam tip: always simplify radicals first and then combine only like surd terms (same radical part).

Simplify each radical: \(\sqrt{20}=2\sqrt{5}\) and \(\sqrt{45}=3\sqrt{5}\). So \(\sqrt{5}+\sqrt{20}-\sqrt{45}=\sqrt{5}+2\sqrt{5}-3\sqrt{5}=(1+2-3)\sqrt{5}=0\). The closest distractor \(-\sqrt{5}\) would arise only from an arithmetic/sign error when combining coefficients; correct combination gives zero. Exam tip: factor out the common \(\sqrt{5}\) first to add coefficients quickly.

The correct answer is \(\sqrt{2}\). Reason: \(\frac{2}{\sqrt{2}}=\frac{2\sqrt{2}}{\sqrt{2}\sqrt{2}}=\frac{2\sqrt{2}}{2}=\sqrt{2}\). Option C, \(\dfrac{\sqrt{2}}{2}\), equals \(\dfrac{1}{\sqrt{2}}\) and is the reciprocal form, so it is incorrect; options B (1) and D (2) are also not equal to the given expression. Exam tip: multiply numerator and denominator by the square root in the denominator to remove the root and simplify quickly.

\(a-1=\sqrt{6}\). Since 6 is not a perfect square, \(\sqrt{6}\) cannot be written as a ratio \(p/q\) of integers, so it is irrational. Option B (rational) is incorrect because \(\sqrt{6}\) is not expressible as a fraction of integers; options C (integer) and D (zero) are incorrect because \(\sqrt{6}\) is neither an integer nor zero. Exam tip: simplify the expression first and check whether the square root is of a perfect square — if not, it is irrational.

Since \(1^2=1<2<4=2^2\), it follows that \(1<\sqrt{2}<2\). Therefore \(\sqrt{2}\) lies between the integers 1 and 2. The other choices are incorrect because they give integers that are too small (0 and 1) or too large (2 and 3; 3 and 4). Exam tip: to locate \(\sqrt{x}\) between integers compare squares of consecutive integers or use a quick decimal estimate (\(\sqrt{2}\approx1.414\)).

For positive numbers a larger number inside the root gives a larger square root. Since (72>50), (\sqrt{72}>\sqrt{50}).

The square root symbol denotes the principal (non-negative) root. Since \(0.81=\frac{81}{100}\), \(\sqrt{0.81}=\sqrt{\frac{81}{100}}=\frac{9}{10}=0.9\). Option B (−0.9) is a common trap because \((-0.9)^2=0.81\) as well, but the principal square root is positive. Options C and D are incorrect: \(0.09^2=0.0081\), and 0.81 is the original number, not its square root. Exam tip: convert decimals to fractions to simplify square-root calculations (e.g., \(0.81=81/100\)).

These are conjugates, so use the difference of squares identity: \((a+b)(a-b)=a^2-b^2\). With \(a=5\) and \(b=\sqrt{2}\), \((5+\sqrt{2})(5-\sqrt{2})=5^2-(\sqrt{2})^2=25-2=23\). Option B (27) is incorrect — it confuses adding instead of subtracting the square; options C and D give irrational forms that are not the product. Exam tip: recognize conjugates and apply the difference of squares to compute such products quickly and reliably.

\(\sqrt[3]{64}=4\) because 64 is a perfect cube (\(4^3\)). The result 4 is an integer, and every integer is rational (can be written as a fraction, e.g. \(4=4/1\)). Option B is incorrect since irrational numbers have non-terminating, non-repeating decimals, which does not apply to 4. Option D is incorrect because a non-repeating decimal implies irrationality, whereas 4 is a terminating decimal. Option C is wrong because non-real numbers involve imaginary parts; \(\sqrt[3]{64}\) is a real number. Exam tip: when evaluating nth roots, first check whether the radicand is a perfect power—if it is, the root is an integer (hence rational).

Prime factorization gives 16 = 2^4. A number is a perfect cube only if every prime exponent is a multiple of 3. Since 4 is not divisible by 3, \(\sqrt[3]{16}\) is not a perfect cube and therefore is irrational. Options B and C are incorrect because 4^3 = 64 and 8^3 = 512, not 16. Option D is wrong because an irrational number cannot be expressed as a terminating decimal. Exam tip: to test whether \(\sqrt[k]{N}\) is rational, factor N and check if every prime exponent is divisible by k.

The key idea is that the square root of a positive integer is rational when the integer is a perfect square, and irrational when it is not a perfect square. Adding the rational number 2 to an irrational number keeps the result irrational. Therefore, we must find the option whose square root cannot be written as a fraction or whole number.

For 18,
\(\sqrt{18}=3\sqrt{2}\), which is irrational because 2 is not a perfect square. In contrast, \(\sqrt{16}=4\), \(\sqrt{25}=5\), and \(\sqrt{36}=6\) are rational. Hence \(2+\sqrt{18}\) is irrational, so option A is correct. The other choices produce rational sums.