Irrational numbers and real numbers

(\sqrt{225}=15), but (226) is not a perfect square. So the first root is rational and the second is irrational.

The digit '2' appears at positions 1, 3, 6, 10, ... which are triangular numbers with general term \(T_n=\tfrac{n(n+1)}{2}\). The gaps between successive '2's increase, so there is no fixed repeating block (period). Any rational number's decimal expansion is either terminating or eventually periodic; this decimal is neither, therefore it is irrational. Exam tip: to test rationality, check whether the decimal eventually repeats or terminates — if not, it is irrational.

The decimal expansion of rational numbers is terminating or repeating. This identification is very useful in exams.

Use the conjugate-product identity:

\((1+\sqrt{3})(1-\sqrt{3})=1^2-(\sqrt{3})^2=1-3=-2.\)

This follows from \((a+b)(a-b)=a^2-b^2\). The tempting wrong choice 4 comes from incorrectly adding 1 and 3 instead of subtracting. Exam tip: apply the conjugate formula and watch the signs carefully when multiplying surds.

(\sqrt{12}=2\sqrt{3}), (\sqrt{75}=5\sqrt{3}), and (\sqrt{27}=3\sqrt{3}). The result is (4\sqrt{3}).

\(\frac{1}{2}\) is rational and \(\sqrt{2}\) is irrational. If their sum were rational, then \(\sqrt{2}=(\frac{1}{2}+\sqrt{2})-\frac{1}{2}\) would be a difference of rationals and thus rational, contradicting that \(\sqrt{2}\) is irrational. Hence \(\frac{1}{2}+\sqrt{2}\) is irrational. The closest distractor "rational" is incorrect for the reason above; "integer" and "terminating decimal" are specific types of rationals so they are also incorrect. Exam tip: memorize that rational + irrational = irrational (contradiction proof is quick and useful in exams).

\(\sqrt{300}=\sqrt{100\times3}=\sqrt{100}\cdot\sqrt{3}=10\sqrt{3}\). Hence option A is correct. The other choices are not equal to \(\sqrt{300}\): \(5\sqrt{6}\approx12.25\) (not 17.32), \(3\sqrt{100}=30\), and \(6\sqrt{50}=6\cdot5\sqrt{2}=30\sqrt{2}\). Exam tip: always extract the largest perfect square factor when simplifying square roots (here 100).

On subtracting, the (4) terms cancel and (\sqrt{7}-(-\sqrt{7})=2\sqrt{7}). Watch the signs carefully.

\(\sqrt{180}=\sqrt{36\times5}=\sqrt{36}\,\sqrt{5}=6\sqrt{5}\), so the simplified form is \(6\sqrt{5}\). Why other options are wrong: \(9\sqrt{2}\) squared is \(81\times2=162\) (not 180), \(5\sqrt{6}\) squared is \(25\times6=150\), and \(3\sqrt{12}=3\sqrt{4\times3}=6\sqrt{3}\), which is different from \(6\sqrt{5}\). Exam tip: always factor out the largest perfect square from under the radical to simplify quickly.

With \(x=\sqrt{17}\), we have \(x^2=(\sqrt{17})^2=17\). Therefore \(x^2-5=17-5=12\). The number 12 is an integer and can be written as \(12/1\), so it is rational. Option B (irrational) is incorrect because irrational numbers cannot be expressed as a ratio of integers, whereas 12 can. Option C (non-real) is wrong because 12 is a real number. Option D (non-repeating decimal) is wrong because 12 is a terminating decimal (12.0), not a non-repeating non-terminating decimal. Exam tip: simplify powers of radicals first — squaring a square root removes the root, then evaluate the remaining arithmetic to classify the result.

A rational number can be written as a fraction of integers, while an irrational number cannot be written in that form. The square root of a positive integer is irrational when the integer is not a perfect square. Since 13 is not a perfect square, \(\sqrt{13}\) is irrational. The number 4, however, is rational because it can be written as \(4/1\).

Adding a rational number to an irrational number always gives an irrational number. If the sum were rational, subtracting the rational number 4 would make \(\sqrt{13}\) rational, which is impossible. Therefore \(4+\sqrt{13}\) is irrational. It is consequently not an integer or a terminating decimal, so option A follows.

(\sqrt{19}) is irrational because (19) is not a perfect square. Subtracting an irrational from a rational gives an irrational result.

Simplify each term: \sqrt{125}=\sqrt{25\times5}=5\sqrt{5} and \sqrt{45}=\sqrt{9\times5}=3\sqrt{5}. Adding like surds gives 5\sqrt{5}+3\sqrt{5}=8\sqrt{5}, so A is correct. Option B (\sqrt{170}) reflects the incorrect assumption \sqrt{a}+\sqrt{b}=\sqrt{a+b}. Option D (10\sqrt{5}) is a mistaken addition of coefficients. Option C (6\sqrt{10}) arises from incorrect factorization or mixing radicands. Exam tip: always factor out perfect squares from each radicand first and then combine only like surds.

Simplify each radical: \(\sqrt{192}=\sqrt{64\times3}=8\sqrt{3}\) and \(\sqrt{27}=\sqrt{9\times3}=3\sqrt{3}\). So the difference is \(8\sqrt{3}-3\sqrt{3}=(8-3)\sqrt{3}=5\sqrt{3}\). The nearest distractor \(3\sqrt{3}\) equals \(\sqrt{27}\), not the difference; a common mistake is to confuse one term with the result. Exam tip: pull out perfect squares first and then combine like surd terms by adding/subtracting their coefficients.

(\sqrt{10}\times\sqrt{40}=\sqrt{400}=20). The product of two irrational numbers can be rational.

(\sqrt{6}\times\sqrt{10}=\sqrt{60}=2\sqrt{15}). Since (15) is not a perfect square the result is irrational.

Add like terms: the rational parts 3 and 3 sum to 6, while the irrational parts \(+\sqrt{6}\) and \(-\sqrt{6}\) cancel. Therefore \(a+b=6\). Distractor D (\(6+2\sqrt{6}\)) is a common mistake resulting from adding the square-root terms instead of canceling them. Exam tip: look for conjugates — the ±√ terms cancel when summed, leaving twice the rational part.

Conjugate multiplication gives ((6)^2-(\sqrt{5})^2=36-5=31). Use (a^2-b^2) in such questions.

This is a product of conjugates; use the identity \((a+b)(a-b)=a^2-b^2\). With \(a=\sqrt{11}\) and \(b=2\), we get \((\sqrt{11})^2-2^2=11-4=7\). Option B (15) is a common mistake from adding 11 and 4 instead of subtracting. Exam tip: recognize conjugates quickly and apply \(a^2-b^2\) to avoid expanding unnecessarily.

Use the square identity \((a+b)^2=a^2+2ab+b^2\). With \(a=3\) and \(b=\sqrt{2}\) we get \((3+\sqrt{2})^2=3^2+2\cdot3\cdot\sqrt{2}+(\sqrt{2})^2=9+6\sqrt{2}+2=11+6\sqrt{2}\). Option B (\(9+6\sqrt{2}\)) omits the \(b^2\) term (+2) and is therefore incorrect. Exam tip: always write all three terms \(a^2,\;2ab,\;b^2\) when expanding a square of a binomial.