Irrational numbers

Step 1: Multiply the given approximate value by (4). Step 2: (4\sqrt{3}\approx4\times1.732=6.928). Step 3: In approximation questions, directly use the given value.

Step 1: (6\sqrt{2}=\sqrt{36}\sqrt{2}). Step 2: This equals (\sqrt{72}). Step 3: To move the outside coefficient inside, multiply by its square.

Step 1: (\sqrt{27}=3\sqrt{3}). Step 2: (x=\sqrt{3}+3\sqrt{3}=4\sqrt{3}). Step 3: Simplify radicals before adding them.

Step 1: (\sqrt{6}\approx2.45) and (\sqrt{10}\approx3.16). Step 2: (3) lies between these two values. Step 3: Use approximate values or squaring to compare.

Step 1: (\sqrt{20}=2\sqrt{5}) and (\sqrt{45}=3\sqrt{5}). Step 2: The total is (6\sqrt{5}), which is irrational. Step 3: Simplify an expression before deciding its nature.

Step 1: Dividing by a non-zero rational number does not remove irrationality. Step 2: For example, (\frac{\sqrt{3}}{4}) remains irrational. Step 3: The condition (k\neq0) is necessary because division by zero is not possible.

Step 1: (\sqrt{216}=6\sqrt{6}) and (\sqrt{54}=3\sqrt{6}). Step 2: (6\sqrt{6}+3\sqrt{6}=9\sqrt{6}). Step 3: Add radicals only after they become like radicals.

Step 1: (4) is a non-zero rational number and (\sqrt{13}) is irrational. Step 2: (4\sqrt{13}) remains irrational. Step 3: Multiplication by zero is a special case, so focus on non-zero rational factors.

Step 1: If (\sqrt{a}=12), square both sides. Step 2: (a=144), and (144) is a perfect square. Step 3: Square both sides in square-root equations to find the original number.

Step 1: (\sqrt{75}=5\sqrt{3}), so (\sqrt{3}+\sqrt{75}=6\sqrt{3}). Step 2: (6\sqrt{3}) is irrational, so it is not rational. Step 3: Simplify each option before deciding its nature.

Step 1: The square root of a perfect square is rational. Step 2: (99) is not a perfect square, so (\sqrt{99}) is irrational. Step 3: In such questions, first separate the perfect squares.

Step 1: (\sqrt{6}) is irrational because (6) is not a perfect square. Step 2: Adding the rational number (5) to an irrational number keeps the result irrational. Step 3: Simplify options before deciding the nature of the number.

Step 1: (\sqrt{147}=7\sqrt{3}) and (\sqrt{75}=5\sqrt{3}). Step 2: (7\sqrt{3}-5\sqrt{3}=2\sqrt{3}). Step 3: Before subtracting radicals, convert them into like radicals.

Step 1: (x-7=(7+\sqrt{13})-7). Step 2: This leaves (\sqrt{13}), which is irrational. Step 3: In such expressions, first subtract the matching rational part.

Step 1: Terminating decimals, fractions, and recurring decimals are rational. Step 2: (\sqrt{63}=3\sqrt{7}), and (\sqrt{7}) is irrational. Step 3: Simplifying the square-root option is a good way to check it.

Step 1: (\sqrt{28}=2\sqrt{7}), (\sqrt{63}=3\sqrt{7}), and (\sqrt{175}=5\sqrt{7}). Step 2: The sum is (2\sqrt{7}+3\sqrt{7}+5\sqrt{7}=10\sqrt{7}). Step 3: Once radicals become like terms, add only the coefficients.

Step 1: Subtracting an irrational number from a rational number leaves an irrational part. Step 2: For example, (5-\sqrt{2}) is irrational. Step 3: For always-type statements, checking counterexamples is useful.

Step 1: First write (\sqrt{9}=3). Step 2: (\frac{9}{\sqrt{9}}=\frac{9}{3}=3). Step 3: Rationalisation is not always needed; first evaluate square roots of perfect squares.

Step 1: Multiplying a square root by itself gives the number inside. Step 2: (\sqrt{k}\times\sqrt{k}=k), which is rational if (k) is an integer. Step 3: The square of an irrational square root can be rational.

Step 1: Since (121<130<144), (11<\sqrt{130}<12). Step 2: (130) is not a perfect square, so (\sqrt{130}) is irrational. Step 3: In interval questions, use nearby perfect squares to set the range.