Irrational numbers and real numbers

(\sqrt{361}=19), but (362) is not a perfect square. So the first root is rational and the second is irrational.

\(\sqrt[3]{343}=7\) since 343 is a perfect cube (\(7^3=343\)) and its cube root is an integer. Every integer is rational (e.g. \(7=7/1\)). Option B (irrational) is incorrect because irrational numbers are non‑terminating, non‑repeating decimals, unlike 7 which is a terminating integer. Option C (non‑real) is incorrect because 7 is a real number. Option D is incorrect because 7 has a terminating decimal representation (7.0), not a non‑terminating non‑repeating decimal. Exam tip: first check if the radicand is a perfect power corresponding to the root; if it is, the root is an integer (hence rational).

20 is not a perfect cube. More formally, if the cube root of an integer n were rational and equal to \(\frac{p}{q}\) in lowest terms, then \(p^3=nq^3\) forces q=1, so the cube root would be an integer. Hence a non–perfect cube has an irrational cube root. Therefore \(\sqrt[3]{20}\) is irrational. The nearby distractor 4 is wrong because \(4^3=64\) (and \(3^3=27\)), so \(\sqrt[3]{20}\approx2.714\) and is not an integer. Exam tip: To test cube roots quickly, check whether the number is one of the perfect cubes 1, 8, 27, 64, ...

(98) is not a perfect square so (\sqrt{98}) is irrational. Adding rational (4) still gives an irrational result.

(\sqrt{225}=15) is rational so (12-\sqrt{225}) will be rational. The root of a perfect square is rational.

Use the identity (a+b)(a-b)=a^2-b^2. Here a=2 and b=\sqrt{5}, so (2+\sqrt{5})(2-\sqrt{5})=2^2-(\sqrt{5})^2=4-5=-1. Option B (1) is a common sign-error distractor; option D (4-\sqrt{5}) still contains the irrational part and is not the simplified product. Exam tip: recognize conjugate pairs and apply a^2-b^2 to eliminate the surd quickly.

The governing concept is the use of conjugate surds and the difference-of-squares identity. Add the two expressions directly: (3+√8)+(3−√8)=3+3+√8−√8=6, because the radical terms cancel. For the product, use (a+b)(a−b)=a²−b². Thus (3+√8)(3−√8)=3²−(√8)²=9−8=1. Therefore option A gives both required values. Option B incorrectly treats the two expressions as opposites, while option C does not cancel the conjugate terms correctly. Option D has the correct sum but wrongly gives the product as 17. The irrationality of √8 does not prevent the product of conjugates from being rational.

Use the square formula \((a-b)^2=a^2+b^2-2ab\). With \(a=\sqrt{7}\) and \(b=\sqrt{3}\) we get \((\sqrt{7}-\sqrt{3})^2=7+3-2\sqrt{7}\sqrt{3}=10-2\sqrt{21}\), so option A is correct. The closest wrong choice \(10+2\sqrt{21}\) has the wrong sign for the \(2ab\) term. Option C \(4-2\sqrt{21}\) arises from confusing signs or subtracting instead of adding \(a^2\) and \(b^2\). Exam tip: always square each term first, then compute the \(-2ab\) term and check its sign before simplifying radicals.

A terminating decimal can be expressed as an integer divided by a power of 10, e.g. as \(\frac{p}{10^n}\). After cancelling common factors this is always of the form \(\frac{a}{b}\) with integers a,b, which is the definition of a rational number. Hence every terminating decimal is rational. Closest distractor: "always an integer" is incorrect because examples like 0.5 or 1.25 are terminating decimals but not integers. Exam tip: convert a terminating decimal to \(\frac{p}{10^n}\) and simplify to quickly show it's rational (or to compare values).

Rational numbers can be expressed as a fraction p/q (integers p and q, q ≠ 0). The decimal expansion of any rational number is either terminating or non‑terminating repeating. Examples: 0.333... = 1/3 and 0.2727... = 27/99 = 3/11. Thus a non‑terminating repeating decimal is rational. Why other options are wrong: B (irrational) is incorrect because irrational numbers have non‑terminating, non‑repeating decimals (e.g. √2). C (non‑real) is incorrect since such decimals represent real numbers. D (integer only) is incorrect because repeating decimals are generally fractions, not necessarily integers. Exam tip: Convert the repeating decimal into a fraction using algebra (e.g., set x = 0.2727..., multiply to shift decimal, subtract) to verify rationality quickly.

\((\frac{2}{5})\) is rational (a ratio of integers) while \(\sqrt{17}\) is irrational because 17 is not a perfect square. If the sum were rational, then \(\sqrt{17}=(\frac{2}{5}+\sqrt{17})-\frac{2}{5}\) would be rational too, which is impossible. Hence the sum is irrational. The closest distractor is "Rational": this would require the irrational part to cancel out, which does not happen here. Options "Integer" and "Terminating decimal" are special cases of rational numbers and are therefore also impossible. Exam tip: isolate the irrational term — subtract any rational parts; if an irrational like \(\sqrt{m}\) with m not a perfect square remains, the whole expression is irrational.

Note that \(\sqrt{275}=\sqrt{25\times11}=5\sqrt{11}\). Therefore \(5\sqrt{11}-\sqrt{275}=5\sqrt{11}-5\sqrt{11}=0\). Option B (\(10\sqrt{11}\)) is incorrect because it is twice the term, option C (\(5\sqrt{11}\)) is just the first term without subtraction, and option D (\(\sqrt{11}\)) is only one-fifth of the term. Exam tip: always simplify radicals by factoring out perfect squares before performing addition or subtraction of surds.

Factor the radicand: \(\sqrt{432}=\sqrt{144\times3}\). Since \(144=12^2\), extract the square root: \(\sqrt{432}=\sqrt{144}\,\sqrt{3}=12\sqrt{3}\). Options B and C are not equal to \(12\sqrt{3}\) numerically; option D simplifies to \(3\sqrt{144}=3\times12=36\), which is incorrect. Exam tip: always look for the largest perfect square factor of the number under the radical and pull its square root outside first.

(\sqrt{242}=11\sqrt{2}), (\sqrt{128}=8\sqrt{2}), and (\sqrt{72}=6\sqrt{2}). The result is (13\sqrt{2}).

This is a difference-of-squares: let a=\sqrt{10}, b=\sqrt{5}. Then (a+b)(a-b)=a^2-b^2. Here a^2=10 and b^2=5, so the value is 10-5=5. Option C (\sqrt{50}) is incorrect because \sqrt{50}=5\sqrt{2}, not 5; option B (1) and D (15) are also wrong — 1 could come from an erroneous division and 15 from adding 10 and 5. Exam tip: Recognise and apply (a+b)(a-b)=a^2-b^2 to simplify quickly without expanding radicals.

Use the identity \((a+b)^2=a^2+2ab+b^2\). Here \(a=2\sqrt{3}\) and \(b=1\). Compute: \((2\sqrt{3})^2=4\cdot3=12,\; 2ab=2\cdot(2\sqrt{3})\cdot1=4\sqrt{3},\; b^2=1\). Summing gives \(12+4\sqrt{3}+1=13+4\sqrt{3}\). Option B (\(12+4\sqrt{3}\)) is the closest wrong choice — it omits the \(+1\). Options C and D have incorrect coefficients for the \(\sqrt{3}\) term or the middle term. Exam tip: Always apply the formula first, simplify each term separately, then add.

Compute square factors: \(\sqrt{27}=\sqrt{9\cdot3}=3\sqrt{3}\) and \(\sqrt{75}=\sqrt{25\cdot3}=5\sqrt{3}\). So \(\sqrt{3}+\sqrt{27}+\sqrt{75}=\sqrt{3}+3\sqrt{3}+5\sqrt{3}=(1+3+5)\sqrt{3}=9\sqrt{3}\). Choice B (\(\sqrt{105}\)) is a common mistake of combining radicals under one root; you can only do that when appropriate (e.g., same radicand). Choices C and D result from incorrect addition of coefficients. Exam tip: always factor out perfect squares from inside radicals first, then combine like radical terms by adding their coefficients.

\(\sqrt{363}=\sqrt{121\times3}=\sqrt{121}\times\sqrt{3}=11\sqrt{3}\). Thus the simplified form is \(11\sqrt{3}\). The closest distractor \(3\sqrt{11}\) is incorrect because \(3\sqrt{11}=\sqrt{9\times11}=\sqrt{99}\), not \(\sqrt{363}\). Options \(\sqrt{33}\) and \(\dfrac{\sqrt{121}}{\sqrt{3}}\) evaluate to different values (the last equals \(11/\sqrt{3}\)). Exam tip: always factor the radicand and pull out the largest perfect square factor to simplify roots quickly.

(\sqrt{243}=9\sqrt{3}), (\sqrt{147}=7\sqrt{3}), and (\sqrt{75}=5\sqrt{3}). The result is (11\sqrt{3}).

The governing concept is comparing square roots by comparing their non-negative radicands, followed by checking whether the result is rational or irrational. Since 10=√100 and 11=√121, and 100<115<121, taking positive square roots gives 10<√115<11. Also, 115 is not a perfect square, so √115 cannot be written as a ratio of integers and is irrational. Thus option A satisfies both conditions. Option B equals √100=10, so it is not strictly between the endpoints. Option C equals √121=11 and is also an endpoint. Option D equals 10.5, which is between 10 and 11 but is rational, not irrational. Therefore only A is valid.