Irrational numbers and real numbers

\(\sqrt{20}=\sqrt{4\times5}=\sqrt{4}\times\sqrt{5}=2\sqrt{5}\). Pull out the largest perfect-square factor (here 4) from under the radical. Option B is not simplified (it’s the original form). Options C and D result from incorrect factor extraction. Exam tip: always factor the radicand and remove the largest perfect square to simplify quickly.

Adding irrational (\sqrt{7}) to rational (3) gives an irrational result. In exams identify roots of non perfect squares.

Option A has the block '27' repeating indefinitely (1.272727...), so it is a non-terminating repeating decimal. Repeating decimals are rational and can be written as fractions. Option C (1.27) is a terminating decimal, not repeating. Option D, \(\sqrt{2}\), is irrational so its decimal expansion is non-terminating and non-repeating. Option B does not show a fixed repeating block in the digits shown, so it is not a repeating decimal (the pattern does not settle into a constant repeating block). Exam tip: to identify a repeating decimal, look for a fixed group of digits that repeats indefinitely after some point.

(\sqrt{14}) is irrational because (14) is not a perfect square. The other options can be written in rational form.

Here \(x^2=(\sqrt{11})^2=11\), so \(x^2+1=12\). The number 12 is an integer and can be written as \(12/1\), therefore it is rational. The closest distractor is "irrational", but irrational numbers cannot be expressed as a ratio of integers whereas 12 can; "non-real" is incorrect because 12 is real; "non-repeating decimal" is wrong because 12 has a terminating decimal form (12.0). Exam tip: when you square a square root, the radical cancels first—compute that before classifying the result.

Simplify the radicals by factoring out square factors: \(\sqrt{75}=\sqrt{25\cdot3}=5\sqrt{3}\) and \(\sqrt{27}=\sqrt{9\cdot3}=3\sqrt{3}\). Subtracting gives \(5\sqrt{3}-3\sqrt{3}=2\sqrt{3}\), so option A is correct. Option B (\(\sqrt{3}\)) is a common distractor arising from subtracting the radical parts incorrectly (treating coefficients as 1 and 0); it equals \(1\sqrt{3}\), not \(2\sqrt{3}\). Exam tip: always factor numbers under the root into square × non-square, simplify each surd, then combine like surds; if unsure, compare decimal values to check your result quickly.

(\sqrt{12}\times\sqrt{3}=\sqrt{36}=6). The product of two irrational numbers is not always irrational.

An irrational decimal must be non-terminating and non-repeating. Option A, 0.3030030003..., shows increasing numbers of zeros between 3s so there is no fixed repeating block; it is non-terminating and non-repeating, hence irrational. Option B (0.303030...) repeats the block “30” and is therefore rational. Options C (0.303) and D (0.3000... which equals 0.3) are terminating decimals and thus rational. Exam tip: check for termination or a fixed repeating pattern — terminating or repeating implies rational; non-terminating non-repeating implies irrational.

Simplify each radical first: \(\sqrt{8}=\sqrt{4\cdot2}=2\sqrt{2}\) and \(\sqrt{18}=\sqrt{9\cdot2}=3\sqrt{2}\). Combine like terms: \(\sqrt{2}+2\sqrt{2}+3\sqrt{2}=(1+2+3)\sqrt{2}=6\sqrt{2}\). Thus the correct answer is \(6\sqrt{2}\). A common wrong choice (e.g. \(4\sqrt{2}\)) comes from mis‑simplifying \(\sqrt{18}\) as \(2\sqrt{2}\); remember \(\sqrt{18}=3\sqrt{2}\). Exam tip: always simplify radicals to the same radicand before adding or subtracting.

Reason: \(\sqrt{12}=\sqrt{4\times3}=\sqrt{4}\times\sqrt{3}=2\sqrt{3}\). Hence option A is correct.
Why others are wrong: Option B (\(\sqrt{12}=4\sqrt{3}\)) and option C (\(\sqrt{12}=3\sqrt{2}\)) are incorrect because the largest perfect square factor of 12 is 4 (not 9 or 16), so only \(\sqrt{4}\) can be taken out. Option D yields a much smaller value and is clearly wrong.
Exam tip: To compare or simplify surds, factor the radicand and pull out the largest perfect square; this makes direct comparison trivial.

\(\sqrt{43}\) is between 6 and 7 because 36 < 43 < 49 implies \(\sqrt{36}=6 < \sqrt{43} < \sqrt{49}=7\). Since 43 is not a perfect square, \(\sqrt{43}\) is irrational. Checking other options: \(\sqrt{36}=6\) equals the lower endpoint, \(\sqrt{50}\approx7.07\) is greater than 7, and \(13/2=6.5\) is rational even though it lies between 6 and 7. Exam tip: compare under-the-root numbers with nearest perfect squares to locate square roots and test for irrationality by checking for perfect squares.

Compute the sum: \(a+b=(2-\sqrt{5})+(2+\sqrt{5})=2+2=4\). Since \(4\) is an integer, it is rational. Option B (irrational) is incorrect because the irrational parts \(-\sqrt{5}\) and \(+\sqrt{5}\) cancel out. Option C (non-real) is wrong because both terms are real numbers so their sum is real. Option D (always negative) is wrong because the sum is +4, not negative. Exam tip: Look for conjugate pairs — the radical parts often cancel when you add conjugates, leaving a rational result.

\(\sqrt{0.25}=0.5\) and \(\sqrt{0.04}=0.2\). Their sum is \(0.5+0.2=0.7\), so 0.7 is correct. Option C (0.54) reflects a common mistake: computing \(\sqrt{0.25+0.04}=\sqrt{0.29}\approx0.5385\) instead of summing the individual square roots. Exam tip: convert decimal squares to fractions (e.g. \(0.25=1/4,\;0.04=1/25\)) to recognize exact square roots quickly.

Solution: \(\dfrac{\sqrt{45}}{\sqrt{5}}=\sqrt{\dfrac{45}{5}}=\sqrt{9}=3\). Use the property \(\dfrac{\sqrt{a}}{\sqrt{b}}=\sqrt{\dfrac{a}{b}}\). Option B (9) reflects a common mistake of removing the radical incorrectly and treating the expression as \(45/5=9\) without taking the square root; note that \(9\neq\sqrt{9}\). Options C and D are wrong because they result from incorrect cancellation or ignoring the denominator. Exam tip: when dividing square roots, combine under a single radical first (simplify \(a/b\)) and then take the square root to avoid algebraic errors.

The overline shows that '12' repeats. Any repeating decimal can be expressed as a ratio of two integers, so it is rational. For example, let \(x=0.\overline{12}\). Then \(100x=12.\overline{12}\); subtracting gives \(99x=12\), hence \(x=12/99=4/33\), a fraction of integers. Choice B is wrong because irrational numbers cannot be written as a ratio of integers; here we have such a ratio. Choice C is wrong because a decimal number is real. Choice D is wrong because the value lies between 0 and 1, so it is not an integer. Exam tip: convert a repeating decimal to a fraction by multiplying to align repeats and subtracting — this quickly shows rationality.

Both \(\sqrt{2}\) and \(-\sqrt{2}\) are irrational, but their sum is \(0\), which is rational. Hence the claim is not always true. Exam tip: test an “always” statement by finding one counterexample.

A rational number has a terminating or recurring decimal expansion. Here the zeros between 1s occur in groups of 1, 2, 3, 4, ... so no fixed repeating block exists. Exam tip: judge by repetition, not merely by the digits used.

Since \(\sqrt{32}=\sqrt{16\times2}=4\sqrt{2}\), we get \(y=\sqrt{2}+4\sqrt{2}=5\sqrt{2}\). Note that \(\sqrt{a}+\sqrt{b}\) cannot be combined as \(\sqrt{a+b}\), so \(\sqrt{34}\) is incorrect. The option \(4\sqrt{2}\) would be the result only if the extra \(\sqrt{2}\) were omitted. Exam tip: always factor the radicand into perfect square factors to extract like radicals before adding or subtracting.

(\sqrt{2}+(-\sqrt{2})=0) is rational but (\sqrt{2}+\sqrt{3}) is irrational. So there is no single fixed rule.

(\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.