Irrational numbers

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}).

The direct answer is A, 162. Simplify the radicals: √32=√(16×2)=4√2 and √50=√(25×2)=5√2. Their sum is 9√2. Squaring gives (9√2)^2=81×2=162, so option A is correct. Option B, 98, is only the sum of the radicands, 32+50, and ignores the cross term created when a sum is squared. Option C, 18√2, is not the square of 9√2; after squaring, the radical should disappear. Option D, 200, comes from an incorrect expansion or multiplication. A second check is 32+50+2√(32×50)=82+2√1600=82+80=162. The reliable method is to simplify each radical, combine like radicals, and then square the complete sum. Never square the two original radicands separately.

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}).

Adding any rational number \(q\) to \(\sqrt{2}\) keeps the result irrational. If \(\sqrt{2}+q\) were rational, subtracting \(q\) would make \(\sqrt{2}\) rational, a contradiction. In \(q\sqrt{2}\), taking \(q=0\) gives 0. Exam tip: rational ± irrational is always irrational.

If \(x+r\) were rational, then \(x=(x+r)-r\) would be rational, which is a contradiction. However, \(x^2\) need not be irrational: \((\sqrt{2})^2=2\). Exam tip: rational ± irrational is always irrational.

A is correct: if \(r+q\) were rational, subtracting \(q\) would make \(r\) rational, a contradiction. For \(r=\sqrt2\), \(r^2=2\), so C fails. Tip: test “always” claims with a counterexample.

Multiplying by the conjugate makes the denominator (14-9=5). So the form is (\frac{5(\sqrt{14}+3)}{5}), that is (\sqrt{14}+3).

Multiplying the same square root by itself gives the number inside. Therefore the value is (8+\sqrt{11}).

Option C is correct. A rational number has a terminating or eventually repeating decimal expansion. Here, the zeros between successive 1s keep increasing, so no fixed repeating block exists. Exam tip: check repetition, not merely the digits used.

Multiplying by the conjugate gives numerator (16+2\sqrt{39}) and denominator (10). So the answer is (\frac{8+\sqrt{39}}{5}).

An irrational number has a non-terminating, non-repeating decimal expansion; no fixed block of digits repeats forever. Terminating and repeating decimals are rational. Exam tip: a recurring digit pattern always indicates a rational number.

\(\frac{\sqrt{45}}{\sqrt{5}}=\sqrt{9}=3\), so it is rational. The other expressions reduce to \(3\sqrt3\), \(\sqrt3\), and \(\sqrt5\), which are irrational. Exam tip: rewrite radicands using perfect-square factors first.

(\frac{7}{\sqrt{23}+4}=\sqrt{23}-4) because the denominator becomes (23-16=7). So the sum is (2\sqrt{23}).

(\sqrt{98}=7\sqrt{2}) and (\sqrt{50}=5\sqrt{2}), so the numerator is (2\sqrt{2}). Dividing gives (2).

If \(x+q\) were rational, then \(x=(x+q)-q\) would be rational, a contradiction. But \(x^2\) need not be irrational: for \(x=\sqrt{2}\), it is 2. Exam tip: use contradiction.

(\sqrt{539}=7\sqrt{11}), (\sqrt{275}=5\sqrt{11}), and (\sqrt{99}=3\sqrt{11}). So the result is (7\sqrt{11}-5\sqrt{11}+3\sqrt{11}=5\sqrt{11}).