Irrational numbers

Step 1: (\sqrt{245}=7\sqrt{5}), (\sqrt{180}=6\sqrt{5}), and (\sqrt{80}=4\sqrt{5}). Step 2: (7\sqrt{5}+6\sqrt{5}-4\sqrt{5}=9\sqrt{5}). Step 3: Before addition or subtraction, write all radicals in like form.

Step 1: This decimal has no fixed block of digits repeating again and again. Step 2: It is non-terminating and non-recurring, so it is irrational. Step 3: While deciding the nature of a decimal, look for a clear repeating rule.

Step 1: (\sqrt{11}) and (-\sqrt{11}) are both irrational. Step 2: Their sum is (0), which is rational. Step 3: Opposite irrational terms can give a rational sum.

Step 1: Apply distribution: (\sqrt{7}\times4+\sqrt{7}\times\sqrt{7}). Step 2: This becomes (4\sqrt{7}+7). Step 3: In such multiplication, remember (\sqrt{7}\times\sqrt{7}=7).

Step 1: (5\sqrt{3}=\sqrt{25}\sqrt{3}). Step 2: This equals (\sqrt{75}). Step 3: When moving an outside coefficient inside the root, multiply by its square.

Step 1: This is of the form \((a+b)(a-b)=a^2-b^2\). Step 2: \(4^2-(\sqrt{7})^2=16-7=9\). Step 3: In conjugate multiplication, directly use the difference of squares.

Step 1: (\sqrt{225}=15). Step 2: (\sqrt{225}-8=7), which is rational. Step 3: When asked for not irrational, carefully check perfect-square options.

Step 1: (ab=(\sqrt{5}+3)(\sqrt{5}-3)). Step 2: Using difference of squares, ((\sqrt{5})^2-3^2=5-9=-4). Step 3: Recognising conjugate form makes the calculation shorter.

Step 1: The conjugate of (4+\sqrt{15}) is (4-\sqrt{15}). Step 2: (\frac{1}{4+\sqrt{15}}\times\frac{4-\sqrt{15}}{4-\sqrt{15}}=\frac{4-\sqrt{15}}{16-15}=4-\sqrt{15}). Step 3: Use the conjugate of the denominator for rationalisation.

Step 1: (\sqrt{3}) and (\sqrt{3}) are both irrational. Step 2: Their product is (3), which is rational. Step 3: Test always-type statements with a counterexample.

Step 1: (\sqrt{44}=2\sqrt{11}), (\sqrt{99}=3\sqrt{11}), and (\sqrt{176}=4\sqrt{11}). Step 2: The sum should be (9\sqrt{11}); the listed options do not contain it. Step 3: If options miss the correct value, the question should be revised.

Step 1: (x^2=(\sqrt{7})^2=7). Step 2: (4x=4\sqrt{7}), so (x^2+4x=7+4\sqrt{7}). Step 3: Simplify the square and the multiplication separately.

Step 1: (12<\sqrt{n}<13) means (144<n<169). Step 2: (150) lies in this interval and is not a perfect square, so (\sqrt{150}) is irrational. Step 3: Square both bounds to handle square-root ranges.

Step 1: A recurring decimal is rational. Step 2: In (0.818181\ldots), the block (81) 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: \((\sqrt{11})^2+2\sqrt{11}\times2+2^2=11+4\sqrt{11}+4=15+4\sqrt{11}\). Step 3: Forgetting the middle term (2ab) is a common mistake.

Step 1: (7<\sqrt{t}<8) means (49<t<64). Step 2: (57) lies in this interval, so (\sqrt{57}) lies between (7) and (8). Step 3: Square the boundary numbers to understand square-root intervals.

Step 1: Multiply numerator and denominator by (\sqrt{3}). Step 2: (\frac{6}{\sqrt{3}}=\frac{6\sqrt{3}}{3}=2\sqrt{3}). Step 3: After rationalising, simplify the answer completely.

Step 1: First multiply the numbers inside the roots. Step 2: The first three give (400), (900), and (784), which are perfect squares; the fourth gives (\sqrt{30}). Step 3: After multiplication, check whether the resulting number is a perfect square.

Step 1: (\sqrt{242}=11\sqrt{2}), (\sqrt{98}=7\sqrt{2}), and (\sqrt{32}=4\sqrt{2}). Step 2: (11\sqrt{2}+7\sqrt{2}-4\sqrt{2}=14\sqrt{2}). Step 3: Convert all radicals into like form before adding or subtracting.

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