Irrational numbers and real numbers

This is a product of conjugates — use the identity \((a+b)(a-b)=a^2-b^2\). With \(a=\sqrt{5}\) and \(b=\sqrt{2}\) we get \(a^2-b^2=5-2=3\). Option C (\(\sqrt{10}\)) confuses the product of the radicals \(\sqrt{5}\cdot\sqrt{2}=\sqrt{10}\) with the whole conjugate product; the middle terms cancel in the conjugate product, so the result is the rational number 3. Option B (\(-3\)) is a sign error. Exam tip: spot conjugates quickly and apply \((a+b)(a-b)=a^2-b^2\) to avoid expanding term-by-term.

First simplify: \(\sqrt{12}=\sqrt{4\times3}=2\sqrt{3}\). So \(2\sqrt{3}+\sqrt{12}=2\sqrt{3}+2\sqrt{3}=4\sqrt{3}\). Option B only gives the first term, option C arises from incorrectly combining under one radical, and option D represents an incorrect arithmetic of coefficients. Exam tip: always extract perfect squares from radicals before adding or subtracting so you combine like surd terms correctly.

\(7.\overline{125}\) is a repeating decimal. Let x = \(7.\overline{125}\). Then 1000x = \(7125.\overline{125}\), so 999x = 7125 − 7 = 7118 and x = 7118/999, a rational number. Option B shows a non‑repeating non‑terminating decimal form, so it's not rational. \(\sqrt{7}\) is irrational because 7 is not a perfect square. \(\pi\) is a well‑known irrational (transcendental) number. Exam tip: look for repeating or terminating decimals (rational) and check whether a square root is of a perfect square (only then rational).

With x = \sqrt{23}, we get \(x^2+2x = 23 + 2\sqrt{23}\). 23 is rational and \(2\sqrt{23}\) is irrational; the sum of a nonzero irrational and a rational number is irrational. Therefore the expression is irrational. The closest distractor, rational, would be correct only if the irrational part canceled to zero, which does not happen here. Exam tip: simplify the expression first and check whether the irrational part can cancel; if not, the result with a nonzero irrational term is irrational.

Since \(\sqrt{12}=2\sqrt{3}\), we have \(\sqrt{3}+\sqrt{12}=\sqrt{3}+2\sqrt{3}=3\sqrt{3}\). Squaring gives \((3\sqrt{3})^2=9\cdot(\sqrt{3})^2=9\cdot3=27\). Choice B (\(9\sqrt{3}\)) is a common mistake: squaring the expression gives a numeric product, not another radical of the same form. Exam tip: combine like surds first, then apply \((a\sqrt{b})^2=a^2b\).

\(\sqrt{85}\) ≈ 9.219…, which lies strictly between 9 and 10. Since 85 is not a perfect square, \(\sqrt{85}\) is irrational. The closest distractor \(\frac{19}{2}=9.5\) is between 9 and 10 but is rational (ratio of integers). \(\sqrt{100}=10\) and \(\sqrt{81}=9\) are endpoints, not numbers strictly between 9 and 10. Exam tip: to find an irrational between n and n+1, pick a non-square integer between n^2 and (n+1)^2 and take its square root — it will be irrational and lie between n and n+1.

\(\frac{25}{4}=6.25\), which lies between 6 and 7 and is rational because it is a ratio of integers. The other choices are irrational: \(\sqrt{41}\) and \(\sqrt{45}=3\sqrt{5}\) are square roots of non-perfect squares, and \(\pi+3\) is irrational because \(\pi\) is transcendental. Exam tip: to find a rational between two integers a and b, use the midpoint \((a+b)/2\) (e.g. \(6.5=13/2\)).

Since \(3^2=9<11<16=4^2\), we have \(3<\sqrt{11}<4\), so \(\sqrt{11}\) lies between 3 and 4 (approximately \(3.316\ldots\)). Option C (4 and 5) is wrong because \(4^2=16>11\), so \(\sqrt{11}\) cannot be ≥4. Exam tip: Compare squares of consecutive integers to bracket a square root quickly on the number line.

For non-negative numbers, the square-root function preserves order: if one number is larger than another, its principal square root is also larger. This follows because squaring non-negative numbers does not reverse their order. Thus comparing the numbers inside the radical signs is enough here.

Since \(145>130\), taking square roots gives \(\sqrt{145}>\sqrt{130}\). The two roots cannot be equal because equal non-negative square roots would have equal squares, which would imply 145 equals 130. Therefore the correct comparison is option A; option B reverses the order, while option C incorrectly treats different radicands as equal.

\sqrt{0.36}=0.6 and \sqrt{0.49}=0.7, so the sum is 0.6+0.7 = 1.3. Option C (1.1) is a tempting close distractor but incorrect because \sqrt{0.49} is 0.7, not 0.5. Exam tip: convert decimals to fractions (e.g. 0.36=(6/10)^2) or memorize common decimal square roots to simplify quickly.

\(\frac{\sqrt{147}}{\sqrt{3}}=\frac{\sqrt{49\cdot3}}{\sqrt{3}}=\frac{7\sqrt{3}}{\sqrt{3}}=7\). Hence the value is 7. The closest distractor is D (\(7\sqrt{3}\)) — that equals \(\sqrt{147}\), not the quotient after dividing by \(\sqrt{3}\). Option B (1) is wrong due to incorrect cancellation of unrelated parts; option C (49) is the result of mistakenly squaring 7. Exam tip: factor out perfect square factors from under the radical first, then simplify by cancelling common radicals.

The governing concept is simplifying a radical denominator by rationalisation. Multiply the numerator and denominator by √2, which is non-zero: 4/√2 = (4√2)/(√2 × √2) = (4√2)/2 = 2√2. This is also consistent with checking the result: (2√2)² = 8, and (4/√2)² = 16/2 = 8; both expressions are positive, so they are equal. Hence option A is correct. Option B is twice the required value. Options C and D each equal 1, whereas 4/√2 is approximately 2.828 and cannot equal 1. The final answer 2√2 has no radical in the denominator and is therefore the preferred simplified exact form.

The square root of a positive integer is rational only when the number is a perfect square. So a non perfect square gives an irrational root.

Distribute: \(\sqrt{2}(5-\sqrt{2})=5\sqrt{2}-\sqrt{2}\cdot\sqrt{2}=5\sqrt{2}-2\). Hence the correct value is \(5\sqrt{2}-2\). The closest distractor \(5-2\sqrt{2}\) is not equal — coefficients and order differ; numerical check: original ≈5.071, distractor ≈2.172. Exam tip: expand using distribution, simplify \(\sqrt{a}\cdot\sqrt{a}=a\), then combine like (rational vs irrational) terms and, if unsure, verify with a quick decimal approximation.

Multiply the coefficients: 3\times2=6. Multiply the radicals: \(\sqrt{6}\times\sqrt{6}=6\). So the product is \(6\times6=36\).
Why the closest distractor is wrong: \(6\sqrt{6}\) results from multiplying only the numerical coefficients (3 and 2) and leaving the radicals uncombined; that ignores \(\sqrt{6}\times\sqrt{6}=6\). 12 and 36\sqrt{6} are also incorrect because they do not follow the correct combination of coefficients and radicals.
Exam tip: Use the identity \(\sqrt{a}\times\sqrt{a}=a\) to simplify products of like square roots quickly.

Combine like radical terms. Each term contains \(\sqrt{3}\), so add their coefficients: \(6-2+1=5\). Thus the simplified form is \(5\sqrt{3}\). Option B (\(9\sqrt{3}\)) reflects mistakenly adding all coefficients as positive (6+2+1) and is incorrect. Option C corresponds to keeping only one unit coefficient; option D wrongly removes the radical. Exam tip: always factor out the common radical (e.g. \(\sqrt{3}\)) and then add/subtract the numeric coefficients before reattaching the radical.

(\sqrt{75}=5\sqrt{3}), (\sqrt{108}=6\sqrt{3}), and (\sqrt{48}=4\sqrt{3}). The result is (7\sqrt{3}), so check option values carefully.

(\sqrt{27}=3\sqrt{3}), (\sqrt{75}=5\sqrt{3}), and (\sqrt{12}=2\sqrt{3}). The result is (6\sqrt{3}).

Let \(x=0.\overline{09}\). Then \(100x=9.\overline{09}\). Subtracting gives \(100x-x=9\Rightarrow 99x=9\), so \(x=9/99=1/11\). Since it equals a fraction of integers, it is rational. "Irrational" is incorrect because irrational numbers have non-terminating, non-repeating decimals; here the decimal repeats. "Non-real" is wrong because every decimal expansion represents a real number, and "Integer" is wrong because the value lies between 0 and 1. Exam tip: convert repeating decimals to fractions by setting the decimal equal to x, multiplying to align repeats, and subtracting to solve for x.

The fractional part \(0.01011011101111\ldots\) shows groups of 1s whose lengths keep increasing (one 1, then two 1s, then three 1s, and so on). There is no fixed block that repeats indefinitely, so the decimal expansion is non-terminating and non-repeating. Hence the number is irrational. Option B is wrong because the decimal does not terminate; C is wrong because a repeating decimal must eventually show a fixed repeating block; D is wrong because the number has a nonzero fractional part. Exam tip: Rational numbers have decimal expansions that either terminate or become eventually periodic — checking for eventual periodicity is a quick test in exams.