Irrational numbers

The governing concept is simplifying radicals by taking perfect-square factors outside the radical and then subtracting like surds. Since 125 = 25×5, √125 = 5√5. Since 45 = 9×5, √45 = 3√5. Thus u = 5√5 − 3√5 = 2√5. Given v = √5, we obtain u − v = 2√5 − √5 = √5. Hence option A is correct. Option B incorrectly treats u and v as equal, option C stops before subtracting v, and option D adds coefficients rather than performing the required subtraction. Once the common radical is identified, only its coefficients need to be operated on.

Multiplying by the conjugate makes the denominator (10-6=4). So (\frac{4(\sqrt{10}+\sqrt{6})}{4}=\sqrt{10}+\sqrt{6}).

(\sqrt{20}=2\sqrt{5}) and (\sqrt{45}=3\sqrt{5}), so the sum is (5\sqrt{5}). Its square is (125).

Both \(\sqrt{18}\) and \(\sqrt{2}\) are irrational, but \(\frac{\sqrt{18}}{\sqrt{2}}=\sqrt{9}=3\), which is rational. Thus the claim is false. Exam tip: one counterexample disproves an “always” statement.

\(\sqrt{75}=5\sqrt{3}\) and \(\sqrt{27}=3\sqrt{3}\), so the bracket is \(2\sqrt{3}\). Multiplying by \(\sqrt{3}\) gives (6).

Multiplying by the conjugate makes the denominator (8-5=3). So (\frac{3(\sqrt{8}-\sqrt{5})}{3}=\sqrt{8}-\sqrt{5}).

The sum of a rational number and an irrational number is always irrational. If their sum were rational, subtracting the rational number would make the irrational number rational, which is impossible. Exam tip: remember this rule for both addition and subtraction.

The governing concept is the square-root property √a×√a=a for a non-negative real number. Here the same quantity, √(4+√7), is multiplied by itself. Therefore the product is [√(4+√7)]²=4+√7. The radicand is positive because √7 is positive, so applying the property is valid. Hence option A is correct. Option B incorrectly squares the two parts separately and then adds them, even though (4+√7)² would include an additional mixed term. Option C is unrelated to the given expression, and option D changes the sign of the radical without justification. Recognising the identical square-root factors avoids unnecessary expansion.

(\sqrt{162}=9\sqrt{2}), (\sqrt{98}=7\sqrt{2}), and (\sqrt{50}=5\sqrt{2}). Therefore the result is (11\sqrt{2}).

The governing idea is rationalisation using the conjugate of a binomial surd. Since t = √7 + 2, we have 3/t = 3/(√7 + 2). Multiply numerator and denominator by √7 − 2: 3(√7 − 2)/[(√7 + 2)(√7 − 2)] = 3(√7 − 2)/(7 − 4) = √7 − 2. Therefore t + 3/t = (√7 + 2) + (√7 − 2) = 2√7. Option A is correct. Option B incorrectly removes the radical, option C fails to cancel the constants, and option D omits one of the equal √7 terms. The conjugate works because it converts the denominator into the rational number 3.

(\sqrt{48}=4\sqrt{3}) and (\sqrt{75}=5\sqrt{3}), so (s=9\sqrt{3}). Dividing by (\sqrt{3}) gives (9).

Multiplying by the conjugate gives (\frac{(\sqrt{5}-\sqrt{2})^2}{5-2}). So the answer is (\frac{7-2\sqrt{10}}{3}).

The square root of a positive integer is rational only when that integer is a perfect square. Hence, if \(n\) is not a perfect square, \(\sqrt{n}\) is irrational. For example, \(12\) is not a perfect square, so \(\sqrt{12}\) is irrational. Exam tip: being even or odd alone is not enough.

\(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4\), which is rational. Since both factors are irrational, this is a counterexample to Reema’s “always” claim. Exam tip: combine square roots first before deciding the type of number.

Area is ((4+\sqrt{5})(4-\sqrt{5})=16-5=11). Multiplying conjugate dimensions can give a rational area.

(\sqrt{300}=10\sqrt{3}), (\sqrt{192}=8\sqrt{3}), and (\sqrt{75}=5\sqrt{3}). Therefore the result is (7\sqrt{3}).

The first term becomes (\frac{\sqrt{5}-\sqrt{3}}{2}) and the second becomes (\frac{\sqrt{5}+\sqrt{3}}{2}). Their sum is (\sqrt{5}).

Both \(\sqrt{2}\) and \(-\sqrt{2}\) are irrational, but \(\sqrt{2}+(-\sqrt{2})=0\), which is rational. Hence Reena’s “always” claim is false. Exam tip: one valid counterexample is enough to disprove an “always” statement.

\(\sqrt{2}\) is irrational, and \(3-\sqrt{2}\) is also irrational; otherwise subtracting it from 3 would make \(\sqrt{2}\) rational. Their sum is \(3\), which is rational. Exam tip: check the sum first in such pairs.

\(3+\sqrt{2}\) is irrational. Dividing it by the non-zero rational number 5 keeps the result irrational. Exam tip: if \(x\) were rational, then \(5x-3=\sqrt{2}\) would be rational, which is a contradiction.