Irrational numbers

Step 1: (\sqrt{18}=3\sqrt{2}), (\sqrt{72}=6\sqrt{2}), and (\sqrt{162}=9\sqrt{2}). Step 2: The sum is (3\sqrt{2}+6\sqrt{2}+9\sqrt{2}=18\sqrt{2}). Step 3: Simplify all radicals completely first.

Step 1: (x^2=(\sqrt{5})^2=5). Step 2: (3x=3\sqrt{5}), so (x^2+3x=5+3\sqrt{5}). Step 3: Simplify the square and the multiplication separately.

Step 1: (10<\sqrt{n}<11) means (100<n<121). Step 2: (105) lies in this interval and is not a perfect square, so (\sqrt{105}) is irrational. Step 3: Square both bounds to handle square-root ranges.

Step 1: A recurring decimal is rational. Step 2: In (0.272727\ldots), the block (27) repeats. Step 3: Identifying the repeating block in a decimal is important.

Step 1: Use \((a+b)^2=a^2+2ab+b^2\). Step 2: \((\sqrt{7})^2+2\sqrt{7}\times1+1^2=7+2\sqrt{7}+1=8+2\sqrt{7}\). Step 3: Forgetting the middle term \(2ab\) is a common mistake.

Step 1: (6<\sqrt{t}<7) means (36<t<49). Step 2: (42) lies in this interval, so (\sqrt{42}) lies between (6) and (7). Step 3: Square the boundary numbers to understand square-root intervals.

Step 1: Multiply numerator and denominator by (\sqrt{2}). Step 2: (\frac{4}{\sqrt{2}}=\frac{4\sqrt{2}}{2}=2\sqrt{2}). Step 3: After rationalising, simplify the answer completely.

Step 1: First multiply the numbers inside the roots. Step 2: The first three give (144), (324), and (900), which are perfect squares; the fourth gives (\sqrt{22}). Step 3: After multiplication, check whether the resulting number is a perfect square.

Step 1: (\sqrt{128}=8\sqrt{2}), (\sqrt{72}=6\sqrt{2}), and (\sqrt{50}=5\sqrt{2}). Step 2: (8\sqrt{2}+6\sqrt{2}-5\sqrt{2}=9\sqrt{2}). Step 3: Convert all radicals into like form before adding or subtracting.

Step 1: Subtracting an irrational number from a rational number leaves an irrational part. Step 2: If the result were rational, then (q=p-(p-q)) would be rational, which is impossible. Step 3: Remember the rules for addition and subtraction of rational and irrational numbers.

Step 1: (\sqrt{c}\times\sqrt{12}=\sqrt{12c}). Step 2: (\sqrt{12c}=18), so (12c=324) and (c=27). Step 3: In square-root equations, square both sides to solve.

Step 1: Since (9<13<16), (\sqrt{13}) lies between (3) and (4). Step 2: (\sqrt{13}\approx3.606), so (3.61) is the closest. Step 3: In approximation, first set the range using perfect squares.

Step 1: (\sqrt{7}) and (\sqrt{7}) are both irrational. Step 2: Their difference is (0), which is rational. Step 3: The difference of equal irrational terms can be rational.

Step 1: (\sqrt{48}\times\sqrt{75}=\sqrt{3600}). Step 2: (\sqrt{3600}=60), so the result is rational. Step 3: In multiplication, multiply the inside numbers and check for a perfect square.

Step 1: In (\frac{1}{3-\sqrt{8}}), the conjugate of the denominator is (3+\sqrt{8}). Step 2: The denominator becomes (9-8=1), so the value is (3+\sqrt{8}). Step 3: Rationalising with the conjugate quickly simplifies the denominator.

Step 1: (6) is rational and (\sqrt{5}) is irrational. Step 2: The sum of a rational and an irrational number is irrational. Step 3: Adding an integer does not remove the irrational square-root part.

Step 1: (\sqrt{363}=11\sqrt{3}) and (\sqrt{75}=5\sqrt{3}). Step 2: (11\sqrt{3}-5\sqrt{3}=6\sqrt{3}). Step 3: Simplify radicals completely before subtracting.

Step 1: (\sqrt{24}) is irrational because (24) is not a perfect square. Step 2: Since (16<24<25), (4<\sqrt{24}<5). Step 3: In condition-based questions, check both irrationality and range.

Step 1: This is of the form \((a-b)(a+b)=a^2-b^2\). Step 2: \((\sqrt{11})^2-(\sqrt{2})^2=11-2=9\). Step 3: In conjugate multiplication, directly use the difference of squares.

Step 1: Rationalise (\frac{1}{2-\sqrt{3}}) by multiplying by (2+\sqrt{3}). Step 2: The denominator becomes (4-3=1), so the value is (2+\sqrt{3}). Step 3: Use rationalisation to identify equivalent forms.