Real Numbers

Both terms are like surds: \(\sqrt{5}\) and \(\sqrt{5}\). Hence, add their coefficients: \(1\sqrt{5}+1\sqrt{5}=2\sqrt{5}\). Therefore, the correct answer is \(2\sqrt{5}\). \(\sqrt{10}\) is incorrect because \(\sqrt{5}+\sqrt{5}\) cannot be written as \(\sqrt{5+5}\). Exam tip: add or subtract only the coefficients of like surds.

For positive numbers, \(\sqrt{a}\times\sqrt{b}=\sqrt{ab}\). Therefore, \(\sqrt{27}\times\sqrt{3}=\sqrt{81}=9\). The number \(27\) is not the square root of \(81\). Exam tip: Combine the radicals first and then look for a perfect square.

\(b=\sqrt{24}=\sqrt{4\times6}=2\sqrt{6}\), while \(a=\sqrt{6}\). Therefore, \(b-a=2\sqrt{6}-\sqrt{6}=\sqrt{6}\). \(2\sqrt{6}\) is the value of \(b\), not the difference. Exam tip: simplify surds to the same radicand before subtracting them.

The set of real numbers is the union of rational and irrational numbers: \(\mathbb{R}=\mathbb{Q}\cup\mathbb{I}\). Rational numbers can be written in the form \(p/q\), whereas irrational numbers cannot be written in this form. Integers, natural numbers, and whole numbers are all subsets of rational numbers, so they do not alone cover all real numbers. Exam tip: remember irrational examples such as \(\sqrt{2}\) and \(\pi\).

For positive numbers, \(\sqrt{a}\times\sqrt{b}=\sqrt{ab}\). Hence, \(\sqrt{12}\times\sqrt{75}=\sqrt{12\times75}=\sqrt{900}=30\). Option 15 is incorrect because \(\sqrt{900}=30\), not 15. Exam tip: Combine surds under one square root and look for a perfect square.

Given \(x=\sqrt{3}+\sqrt{2}\), \(x^2=(\sqrt{3}+\sqrt{2})^2=3+2+2\sqrt{3}\sqrt{2}=5+2\sqrt{6}\). Therefore, \(x^2-5=2\sqrt{6}\). The option \(\sqrt{6}\) misses the factor 2 in the middle term. Exam tip: while using \((a+b)^2=a^2+b^2+2ab\), do not omit \(2ab\).

Option A simplifies to \((-7/5)/(-7/5)=1\). The number 1 is an integer and can be written as a ratio of integers (\(p/q\), \(q\neq0\)), so it is rational. Option B, \(\sqrt{21}\), is not a perfect square and therefore irrational. Option C is \(\sqrt{31}/5\); dividing a nonzero irrational by a nonzero rational gives an irrational number, so C is irrational. Option D contains \(\pi\), which is irrational, so \((\pi+1)/2\) remains irrational. Exam tip: simplify expressions first — identical nonzero numerator and denominator give 1 (rational); check for perfect squares and known irrational constants to identify irrational options quickly.

Since \(147=49\times 3=7^2\times 3\), \(\sqrt{147}=\sqrt{7^2\times3}=7\sqrt{3}\). The option \(3\sqrt{7}\) would correspond to \(\sqrt{9\times7}=\sqrt{63}\), so it is not correct. Exam tip: identify the greatest perfect-square factor before simplifying a surd.

Given \(\sqrt{x}=12\), squaring both sides gives \(x=12^2=144\). Therefore, \(x-44=144-44=100\). The value 144 is the value of \(x\), not of the required expression \(x-44\). Exam tip: First square a square-root equation to find the variable, then substitute it into the expression asked.

Irrational numbers have non-terminating, non-repeating decimal expansions. The square root of a perfect square is rational (for example \(\sqrt{4}=2\)), while the square root of a non-perfect square is irrational. In option A, \(\sqrt{6},\ \sqrt{10},\ \sqrt{15}\) are not square roots of perfect squares, so all three are irrational. The closest distractor B is wrong because \(\sqrt{4}=2\) is rational. Options C and D are also incorrect since \(\tfrac{1}{2}\) and \(0\) are rational. Exam tip: when checking square roots, first test if the radicand is a perfect square — if it is, the root is rational.

Since \(32=16\times2\), \(\sqrt{32}=\sqrt{16\times2}=4\sqrt{2}\). Similarly, \(8=4\times2\), so \(\sqrt{8}=2\sqrt{2}\). Therefore, \(4\sqrt{2}+2\sqrt{2}=6\sqrt{2}\), making option B correct. \(4\sqrt{2}\) is only the value of \(\sqrt{32}\), not their sum. Exam tip: simplify surds into like radical terms, such as \(\sqrt{2}\), before adding them.

Given \(a=\sqrt{10}\), squaring gives \(a^2=(\sqrt{10})^2=10\). Therefore, \(\frac{a^2}{2}=\frac{10}{2}=5\). The value \(10\) is only \(a^2\); it still needs to be divided by 2. Exam tip: remember that \((\sqrt{x})^2=x\).

Since \(108=36\times3\), and \(36\) is a perfect square, \(\sqrt{108}=\sqrt{36\times3}=\sqrt{36}\sqrt{3}=6\sqrt{3}\). The option \(3\sqrt{6}\) is not correct because its square is \(54\), not \(108\). Exam tip: To simplify a surd, first identify the greatest perfect-square factor of the number.

Given \(x=\sqrt{2}\), \(3x-x=(3-1)x=2x=2\sqrt{2}\). The option \(\sqrt{2}\) is only the value of \(x\), whereas the expression asks for \(2x\). Exam tip: when adding or subtracting like terms, add or subtract only their coefficients.

Since \(63=9\times7\) and \(28=4\times7\), \(\sqrt{63}=3\sqrt{7}\) and \(\sqrt{28}=2\sqrt{7}\). Therefore, \(\sqrt{63}-\sqrt{28}=3\sqrt{7}-2\sqrt{7}=\sqrt{7}\). \(2\sqrt{7}\) is only the simplified form of \(\sqrt{28}\), not the difference. Exam tip: simplify surds into like terms before subtracting them.

Given \(x=\sqrt{15}\), we get \(x^2=(\sqrt{15})^2=15\). Therefore, \(x^2+10=15+10=25\), so option C is correct. Option 20 would result from adding 5 instead of 10. Exam tip: squaring a square root gives the original non-negative number.

\(\sqrt{49}=7\), and 7 is a real number, so option C is correct. Options A, B, and D contain square roots of negative numbers, which are not defined in the real number system. Exam tip: A square root is real only when the number inside the radical is zero or positive.

Since \(300=100\times3=10^2\times3\), \(\sqrt{300}=\sqrt{10^2\times3}=10\sqrt{3}\). The close distractor \(5\sqrt{3}\) squares to \(75\), not \(300\). Exam tip: to simplify a surd, first identify the greatest perfect-square factor of the number.

\(\sqrt{8}=\sqrt{4\times2}=2\sqrt{2}\). Therefore, \(a=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\), so option B is correct. \(2\sqrt{2}\) is only the simplified form of \(\sqrt{8}\); the remaining \(\sqrt{2}\) must still be added. Exam tip: simplify surds by identifying perfect-square factors inside the radical.

Rational numbers can be written as fractions, integers, terminating decimals or repeating decimals. In option A, \(0.125\) is a terminating decimal (\(\tfrac{1}{8}\)), \(-4\) is an integer, and \(\sqrt{169}=13\) is an integer — so all are rational. A common distractor is D: \(\sqrt{18}=3\sqrt{2}\) is irrational, so D is not all rational. Exam tip: check whether a square root is of a perfect square and whether decimals terminate or repeat to decide rationality quickly.