Irrational numbers

(\frac{1}{\sqrt{17}+4}=\sqrt{17}-4) because the denominator becomes (17-16=1). So the sum is (2\sqrt{17}).

An irrational number has a non-terminating, non-repeating decimal expansion, so A is correct. B fails since \((\sqrt{2})^2=2\) is rational. Exam tip: repeating decimals are rational.

The square of a square root gives the number inside. So \(\left(\sqrt{10+\sqrt{21}}\right)^2=10+\sqrt{21}\).

Area is ((\sqrt{14}+\sqrt{5})(\sqrt{14}-\sqrt{5})=14-5=9). Conjugate dimensions can give a rational area.

The governing concept is simplifying surds by extracting perfect-square factors and then combining like surds. Since 8=4×2, √8=2√2. Since 18=9×2, √18=3√2. Therefore y=2√2+3√2=5√2. Dividing by √2 gives y/√2=(5√2)/√2=5, because √2 is non-zero. Hence option C is correct. Option A or B may result from simplifying only one radical or losing a coefficient, while option D can arise from adding coefficients incorrectly. The radicands 8 and 18 must not be added directly; both radicals first need to be expressed as multiples of the same irrational factor √2.

An irrational number has an endless decimal expansion with no repeating block. In contrast, \(1/8=0.125\) terminates, so it is rational. Exam tip: choose non-terminating, non-repeating decimals.

\(\sqrt{12}\) and \(2-\sqrt{12}\) are irrational; otherwise subtracting from 2 would make \(\sqrt{12}\) rational. Their sum is \(2\), which is rational. Option A gives \(5\sqrt{3}\). Check both conditions in such questions.

\(x+r\) is irrational: if it were rational, subtracting the rational number \(r\) would make \(x\) rational, a contradiction. Also, \(x/x=1\) is rational. Exam tip: adding a rational number does not change irrationality.

Since (7<7.5<8), (\sqrt{7.5}) lies between them. Compare square roots using the numbers inside.

\(\sqrt{3}\) is irrational because 3 is not a perfect square, yet \(\sqrt{3}\times\sqrt{3}=3\), a rational number. Option B still gives irrational \(\sqrt6\). Exam tip: one valid counterexample disproves an “always” statement.

Since \(m\) is not a perfect square, \(\sqrt{m}\) is irrational. A rational \(a+b\sqrt{m}\) would give \((a+b\sqrt{m}-a)/b=\sqrt{m}\) rational, a contradiction. Tip: check \(b\ne0\).

(\sqrt{300}=10\sqrt{3}), (\sqrt{108}=6\sqrt{3}), and (\sqrt{75}=5\sqrt{3}). Therefore the result is (9\sqrt{3}).

If \(rx\) were rational, then \(x=\frac{rx}{r}\) would also be rational because \(r\neq0\) is rational. This is a contradiction. However, \(x^2\) can be rational, for example when \(x=\sqrt{2}\). Exam tip: multiplying by a non-zero rational preserves irrationality.

(\sqrt{72}=6\sqrt{2}) and (\sqrt{128}=8\sqrt{2}), so the numerator is (14\sqrt{2}). Dividing gives (14).

(r^2-s^2=(r-s)(r+s)) where (r-s=2\sqrt{2}) and (r+s=2\sqrt{7}). So the value is (4\sqrt{14}).

(\sqrt{242}=11\sqrt{2}), (\sqrt{50}=5\sqrt{2}), and (\sqrt{98}=7\sqrt{2}). Therefore (P=9\sqrt{2}).

(\sqrt{28}=2\sqrt{7}) and (\sqrt{63}=3\sqrt{7}), so the sum is (5\sqrt{7}). Its square is (175).

If \(x+r\) were rational, subtracting the rational number \(r\) would make \(x\) rational, which is a contradiction. \(x^2\) is not always irrational; for \(x=\sqrt{2}\), it equals 2. Exam tip: rational ± irrational is irrational.

If \(x+3\) were rational, subtracting the rational number 3 would make \(x\) rational, a contradiction. However, \(x^2\) can be rational, for example when \(x=\sqrt{2}\). Exam tip: rational ± irrational is always irrational.

The governing concept is expressing all radicals with a common square-free factor before adding and dividing. We have √18=√(9×2)=3√2, √50=√(25×2)=5√2, and √8=√(4×2)=2√2. Thus the numerator is 3√2+5√2=8√2. The entire expression becomes (8√2)/(2√2)=8/2=4, because √2 is a common nonzero factor. Therefore option B is correct. Option A may result from using only part of the numerator. Option C can arise from adding radicands or coefficients incorrectly. Option D ignores the denominator. The essential rule is that radicals should first be reduced to like surds; only then can their coefficients be combined and the common factor cancelled safely.