Real Numbers

Since \(72=36\times2\) and \(36\) is a perfect square, \(\sqrt{72}=\sqrt{36\times2}=\sqrt{36}\times\sqrt{2}=6\sqrt{2}\). Therefore, option B is correct. The coefficients in options A and D are incorrect, while option C does not square to \(72\). Exam tip: factor the number using the largest perfect-square factor before simplifying a square root.

Since \(63=9\times7\) and \(9\) is a perfect square, \(\sqrt{63}=\sqrt{9\times7}=\sqrt{9}\times\sqrt{7}=3\sqrt{7}\). Option A results from incorrectly treating \(7\) as a perfect square. In exams, identify the greatest perfect-square factor of the number under the radical.

The governing concept is the product rule for square roots: for non-negative numbers, √a × √b = √(ab). Applying it gives √2 × √50 = √(2 × 50) = √100 = 10, because the principal square root of 100 is 10. Therefore option D is correct. The result can also be checked by simplifying √50 first: √50 = √(25 × 2) = 5√2. Hence √2 × √50 = √2 × 5√2 = 5(√2)² = 5 × 2 = 10. Option A incorrectly treats the factors as if their square roots produced 25. Option B is only an intermediate expression before multiplying the remaining surds, so it is not the final value. Option C incorrectly adds the radicands instead of multiplying them.

The quotient of square roots can be simplified by using \\(\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\\) when the denominator is positive. Here, both 80 and 5 are positive, so the rule applies directly. Dividing the numbers inside the radical gives \\(\sqrt{80}\div\sqrt{5}=\sqrt{\frac{80}{5}}=\sqrt{16}\\).

The square root of 16 is 4 because \\(4\times4=16\\). Therefore, option A is correct. Another valid route is to write \\(\sqrt{80}=\sqrt{16\times5}=4\sqrt{5}\\), and then divide by \\(\sqrt{5}\\), again obtaining 4. The other numerical choices result from incorrect treatment of the square roots.

Both terms contain the same surd, \(\sqrt{2}\), so their coefficients can be added: \(4\sqrt{2}+5\sqrt{2}=(4+5)\sqrt{2}=9\sqrt{2}\). Option A uses an incorrect sum of coefficients. Option C equals \(9\sqrt{4}=18\), while option D simplifies to \(\sqrt{18}=3\sqrt{2}\), so neither is correct. Exam tip: coefficients can be added or subtracted only when the surd parts are identical.

An infinite decimal can still be rational if it repeats. Here, the number of zeros between successive 1s keeps increasing, so no fixed repeating block exists. Exam tip: a non-terminating, non-repeating decimal is irrational.

\(\sqrt{48}=\sqrt{16\times3}=4\sqrt{3}\) and \(\sqrt{27}=\sqrt{9\times3}=3\sqrt{3}\). Hence, \(\sqrt{48}+\sqrt{27}=4\sqrt{3}+3\sqrt{3}=7\sqrt{3}\), so option A is correct. Option B incorrectly treats the sum of two square roots as \(\sqrt{48+27}=\sqrt{75}\). Exam tip: simplify each radical first, then combine terms having the same radical part.

\(0.375=\frac{375}{1000}=\frac{3}{8}\), so it can be written as \(p/q\) and is rational. Option B is wrong because terminating decimals are also rational. Exam tip: terminating or recurring decimals are rational.

To rationalise the denominator, multiply the numerator and denominator by \(\sqrt{2}\): \(\frac{3}{\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}=\frac{3\sqrt{2}}{2}\). Therefore, option B is correct. In option A, the factor \(\sqrt{2}\) is incorrectly omitted from the numerator. Exam tip: When the denominator contains \(\sqrt{a}\), multiply both numerator and denominator by \(\sqrt{a}\).

Multiplying numerator and denominator by ( \sqrt{7} ) gives ( \frac{5\sqrt{7}}{7} ). Rationalising the denominator is a common exam question.

To rationalise a denominator containing a square root, multiply the fraction by the conjugate of the denominator. The conjugate of \\(2+\sqrt{3}\\) is \\(2-\sqrt{3}\\). Multiplying by \\(\frac{2-\sqrt{3}}{2-\sqrt{3}}\\) does not change the value, but it changes the denominator into a difference of squares.

Thus, \\(\frac{1}{2+\sqrt{3}}\times\frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{2^2-(\sqrt{3})^2}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}\\). The resulting denominator is 1, so it is rational. Therefore, option D is correct. The conjugate is essential; multiplying by the same expression would not remove the radical from the denominator.

The conjugate of \(3-\sqrt{5}\) is \(3+\sqrt{5}\). Therefore, multiply the numerator and denominator by \(3+\sqrt{5}\): \(\frac{1}{3-\sqrt{5}}\times\frac{3+\sqrt{5}}{3+\sqrt{5}}=\frac{3+\sqrt{5}}{9-5}=\frac{3+\sqrt{5}}{4}\). Hence, the new numerator is \(3+\sqrt{5}\). Remember that the conjugate is formed by changing only the sign between the two terms; \(8\) is not the denominator, since \(9-5=4\).

The two factors are conjugates, so the difference-of-squares identity (a+b)(a−b) = a²−b² applies directly. Here a = 2 and b = √5. Thus (2+√5)(2−√5) = 2² − (√5)² = 4 − 5 = −1, making option B correct. Direct expansion confirms the result: 4 − 2√5 + 2√5 − 5. The middle terms cancel because they have equal magnitudes and opposite signs, leaving −1. Option A could arise from using the wrong sign between the squares. Option C ignores the fact that 2² is 4, and option D fails to simplify the conjugate product. The cancellation of the radical term is the key structural idea.

\(\pi\) is irrational because it cannot be expressed as a ratio \(\frac{p}{q}\) of two integers with \(q\neq 0\). In contrast, \(0.25=\frac{1}{4}\), \(\sqrt{49}=7\), and \(-7\) are all rational numbers. Exam tip: the square root of a perfect square is rational, whereas \(\pi\) and, generally, the square root of a non-perfect square are irrational.

Since (12>3), ( \sqrt{12}>\sqrt{3} ). For positive numbers, a larger radicand gives a larger square root.

( \sqrt{10}\approx3.16 ), so ( -\sqrt{10} ) is less than ( -3 ). The negative number closer to zero is greater.

Since \(1.2^2=1.2\times1.2=1.44\), we have \(\sqrt{1.44}=1.2\). The principal square root is always taken as non-negative. The other options are incorrect because \(0.12^2=0.0144\), \(1.02^2=1.0404\), and \(12^2=144\). As an exam tip, square the proposed answer to check the decimal placement.

A rational number can be written as \(\frac{p}{q}\), where p and q are integers and \(q\ne0\). For example, \(-7=\frac{-7}{1}\), so every integer is rational. Non-terminating, non-recurring decimals are irrational. Exam tip: write an integer over 1 to test it.

Using the square-root rule, \(\sqrt{\frac{25}{36}}=\frac{\sqrt{25}}{\sqrt{36}}=\frac{5}{6}\). Therefore, option C is correct. Option B, \(\frac{6}{5}\), is the reciprocal of the required value, not the value itself. Exam tip: the principal square root is taken as the non-negative value.

Using the property of square roots, \(\sqrt{\frac{121}{144}}=\frac{\sqrt{121}}{\sqrt{144}}\). Since \(\sqrt{121}=11\) and \(\sqrt{144}=12\), the value is \(\frac{11}{12}\). Exam tip: For a positive fraction, take the square root of the numerator and denominator separately.