Irrational numbers

(\sqrt{242}=11\sqrt{2}), (\sqrt{128}=8\sqrt{2}), and (\sqrt{72}=6\sqrt{2}). Therefore the result is (9\sqrt{2}).

In the first option, both terms are irrational but they are additive inverses. Their sum is 0, which is rational. Hence, the sum of two irrational numbers need not be irrational. Exam tip: test an “always” statement by looking for one counterexample.

Option B is correct. If \(r+\sqrt{2}\) were rational, subtracting the rational number \(r\) would make \(\sqrt{2}\) rational, which is impossible. Exam tip: rational ± irrational is always irrational.

Since 7 is not a perfect square, \(\sqrt{7}\) is irrational. If the sum were rational, subtracting rational 5 would make \(\sqrt{7}\) rational, a contradiction. Exam tip: use closure under subtraction.

Suppose \(rx\) were rational. Then \(x=(rx)/r\) would also be rational because \(r\ne0\), which is a contradiction. Hence \(rx\) is irrational. Since \(\sqrt2\cdot\sqrt2=2\), option D is not always true. Exam tip: test every “always” statement using a counterexample.

(\sqrt{363}=11\sqrt{3}), (\sqrt{147}=7\sqrt{3}), and (\sqrt{75}=5\sqrt{3}). Therefore (P=13\sqrt{3}).

The denominator \(\sqrt{26}+\sqrt{17}\) contains two square roots, so its conjugate \(\sqrt{26}-\sqrt{17}\) is used. This changes the denominator into a difference of squares and makes it rational. Multiplying by the conjugate in both numerator and denominator keeps the fraction equal to its original value.

The denominator becomes \((\sqrt{26}+\sqrt{17})(\sqrt{26}-\sqrt{17})=26-17=9\). Hence \(\frac{9}{\sqrt{26}+\sqrt{17}}=\frac{9(\sqrt{26}-\sqrt{17})}{9}=\sqrt{26}-\sqrt{17}\). Therefore option A is correct. The cancellation of the factor 9 is valid because the denominator difference is exactly 9.

\(p+q\) must be irrational. If it were rational, then \(q=(p+q)-p\) would be the difference of two rational numbers and hence rational, a contradiction. But \(q^2\) need not be irrational: \((\sqrt{2})^2=2\). Exam tip: adding a rational number to an irrational number gives an irrational result.

The direct answer is A: \\(\\frac{17+2\\sqrt{66}}{5}\\). To rationalise the denominator, multiply numerator and denominator by the conjugate \\(\\sqrt{11}+\\sqrt{6}\\). The denominator becomes \\( (\\sqrt{11}-\\sqrt{6})(\\sqrt{11}+\\sqrt{6})=11-6=5\\). The numerator becomes \\( (\\sqrt{11}+\\sqrt{6})^2=11+6+2\\sqrt{66}=17+2\\sqrt{66}\\). Hence the simplified result is \\(\\frac{17+2\\sqrt{66}}{5}\\). Option A is exactly this result. Option B has the wrong sign in the numerator; the cross term is positive because both terms in the numerator are added. Option C omits the denominator 5, so it is too large. Option D is simply 1, but the original fraction is not 1; its numerator and denominator are different. The useful rule is: multiply by the conjugate, use difference of squares in the denominator, then expand the square in the numerator.

\(2-\sqrt{2}\) is irrational, yet \(\sqrt{2}+(2-\sqrt{2})=2\), which is rational. Hence the statement is false. Exam tip: test “always” claims using a counterexample.

(\sqrt{605}=11\sqrt{5}), (\sqrt{320}=8\sqrt{5}), and (\sqrt{125}=5\sqrt{5}). Therefore the result is (8\sqrt{5}).

\(\sqrt{5}\) is irrational, but \((\sqrt{5})^2=5\), which is rational. Thus, a rational square does not guarantee a rational number. In exams, check the nature of both the number and its square.

(\sqrt{20}=2\sqrt{5}), (\sqrt{45}=3\sqrt{5}), and (\sqrt{80}=4\sqrt{5}). So (y=9\sqrt{5}) and (y^2=405).

To remove the irrational denominator \(\sqrt{30}-\sqrt{23}\), multiply by its conjugate \(\sqrt{30}+\sqrt{23}\). A conjugate changes only the sign between the two terms. Their product is a difference of squares, so the radicals disappear from the denominator.

The denominator becomes \((\sqrt{30}-\sqrt{23})(\sqrt{30}+\sqrt{23})=30-23=7\). Therefore \(\frac{7}{\sqrt{30}-\sqrt{23}}=\frac{7(\sqrt{30}+\sqrt{23})}{7}=\sqrt{30}+\sqrt{23}\). This is the expression represented by option C. The other choices do not simplify to the required result or retain unnecessary factors.

The direct answer is C, \\(15+\\sqrt{26}\\). Let the positive square-root expression be \\(x=\\sqrt{15+\\sqrt{26}}\\). The question asks for \\(x\\times x\\), which is \\(x^2\\). By the meaning of a square root, \\(\\left(\\sqrt{a}\\right)^2=a\\) when the radicand is non-negative. Here the radicand is \\(15+\\sqrt{26}\\), which is positive, so the result is exactly \\(15+\\sqrt{26}\\). Option A, 225+26, incorrectly squares the two parts separately. Option B, \\(\\sqrt{41}\\), changes the expression without a valid identity. Option C is correct. Option D, \\(15-\\sqrt{26}\\), changes the plus sign to minus and is unsupported. Memory cue: the same square root multiplied by itself returns its radicand.

In option C, the decimal continues forever without repeating a fixed block; the number of zeros keeps increasing. Hence it is irrational. A terminates and B repeats, so both are rational. Exam tip: a non-terminating, non-repeating decimal represents an irrational number.

Multiplying by the conjugate gives numerator (22+2\sqrt{85}) and denominator (12). The simplified form is (\frac{11+\sqrt{85}}{6}).

\(x-x=0\), and \(0=0/1\), so it is rational. The expressions \(x+1\), \(3x\), and \(x/2\) remain irrational. Exam tip: identify exact cancellation of like terms quickly.

If \(p+x\) were rational, then \(x=(p+x)-p\) would also be rational, which is a contradiction. Hence A is correct; B, C and D wrongly call the result rational. Exam tip: use contradiction for such properties.

(\frac{6}{\sqrt{31}+5}=\sqrt{31}-5) because the denominator becomes (31-25=6). So the sum is (2\sqrt{31}).

(\sqrt{300}=10\sqrt{3}) and (\sqrt{108}=6\sqrt{3}), so the numerator is (4\sqrt{3}). Dividing gives (4).

Assume \(rx\) is rational. Since \(r\ne0\) is rational, \(x=(rx)/r\) would also be rational, giving a contradiction. Hence \(rx\) is irrational. In exams, always check that the rational multiplier is non-zero.

(\sqrt{847}=11\sqrt{7}), (\sqrt{363}=7\sqrt{7}), and (\sqrt{147}=3\sqrt{7}). So the result is (11\sqrt{7}-7\sqrt{7}+3\sqrt{7}=7\sqrt{7}).

Multiplying by the conjugate makes the denominator (27-11=16). So the form is (\frac{8(\sqrt{27}+\sqrt{11})}{16}).

In A, \(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4\), which is rational, so the claim fails. In B, \(\sqrt{6}\) remains irrational. Exam tip: test “always” using one counterexample.