Irrational numbers

Both \(\sqrt{2}\) and \(\sqrt{8}\) are irrational, but their product is \(\sqrt{16}=4\), a rational number. Thus the word “always” makes the claim false. Exam tip: one valid counterexample is enough to disprove a universal statement.

Multiplying by the conjugate makes the denominator (17-8=9). So the form is (\frac{5(\sqrt{17}-\sqrt{8})}{9}).

((\sqrt{14}-\sqrt{7})^2=14+7-2\sqrt{98}). Since (\sqrt{98}=7\sqrt{2}), the answer is (21-14\sqrt{2}).

The governing concepts are the area formula for a rectangle and the difference-of-squares identity. The area is length multiplied by breadth, so A = (√18 + √2)(√18 − √2). Let a = √18 and b = √2. Using (a + b)(a − b) = a² − b², we obtain A = (√18)² − (√2)² = 18 − 2 = 16. Therefore, option B is correct. The irrational-looking dimensions form a conjugate pair, so the radical terms cancel in the product and the area becomes the rational number 16. Option A incorrectly adds 18 and 2. Options C and D retain radicals even though the identity removes them. Both dimensions are positive, so the result is a valid positive area.

The governing concept is reducing all radicals to like surds before combining coefficients. Since 98=49×2, √98=7√2. Since 50=25×2, √50=5√2. Substitution gives u=3√2+7√2−5√2=(3+7−5)√2=5√2. Therefore u/√2=(5√2)/√2=5, as √2 is non-zero. Option C is correct. The subtraction sign must remain attached to the coefficient of √50. Options A, B and D may result from simplifying √98 or √50 incorrectly, ignoring the minus sign, or combining the original radicands rather than their coefficients. Writing every term as a multiple of √2 makes the calculation clear and reliable.

(\sqrt{45}=3\sqrt{5}) and (\sqrt{80}=4\sqrt{5}), so the numerator is (7\sqrt{5}). Dividing by (\sqrt{5}) gives (7).

\(x^2\) need not be irrational. For example, \(x=\sqrt{2}\) is irrational, but \(x^2=2\) is rational. Adding a rational number to, or multiplying/dividing an irrational number by a non-zero rational, keeps it irrational. Exam tip: test such claims using \(\sqrt{2}\).

Option A is correct. If \(qx\) were rational, then \(x=(qx)/q\) would also be rational because \(q\ne0\) is rational. This contradicts that \(x\) is irrational. Exam tip: always check the non-zero condition.

The square of a square root gives the number inside. So \(\left(\sqrt{12+\sqrt{35}}\right)^2=12+\sqrt{35}\).

If \(r+q\) were rational, then \(r=(r+q)-q\) would also be rational, a contradiction. Hence adding a rational number to an irrational number remains irrational. For \(r=\sqrt2\), \(r^2=2\). Exam tip: use contradiction for such properties.

Multiplying by the conjugate makes the denominator (19-12=7). So (\frac{7(\sqrt{19}+\sqrt{12})}{7}=\sqrt{19}+\sqrt{12}).

(\sqrt{363}=11\sqrt{3}), (\sqrt{147}=7\sqrt{3}), and (\sqrt{75}=5\sqrt{3}). Therefore (P=9\sqrt{3}).

(\frac{10}{\sqrt{19}+3}=\sqrt{19}-3) because the denominator becomes (19-9=10). So the sum is (2\sqrt{19}).

(\sqrt{32}=4\sqrt{2}) and (\sqrt{50}=5\sqrt{2}), so the sum is (9\sqrt{2}). Its square is (162).

Both \(\sqrt{2}\) and \(\sqrt{8}\) are irrational, but \(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4\), which is rational. Hence the statement is false. In exams, disprove “always” statements using one counterexample.

Adding both fractions gives numerator (2\sqrt{12}) and denominator (12-5=7). So the value is (\frac{4\sqrt{3}}{7}).

(a^2-b^2=(a-b)(a+b)), where (a-b=2\sqrt{8}) and (a+b=2\sqrt{18}). So the value is (4\sqrt{144}=48).

(\sqrt{243}=9\sqrt{3}) and (\sqrt{108}=6\sqrt{3}), so the numerator is (3\sqrt{3}). Dividing by (\sqrt{3}) gives (3).

If \(3+\sqrt{2}\) were rational, subtracting the rational number 3 would make \(\sqrt{2}\) rational. But \(\sqrt{2}\) is irrational, so the claim is false. Exam tip: use contradiction for such proofs.

Multiplying by the conjugate gives numerator (9+2\sqrt{14}) and denominator (5). So the simplified form is (\frac{9+2\sqrt{14}}{5}).