Irrational numbers and real numbers

On adding the two squares the radical terms cancel and the result is (32). Identify cancelling terms in conjugates.

By identity the difference is (4ab), where (a=\sqrt{11}) and (b=\sqrt{5}). So the answer is (4\sqrt{55}).

(\sqrt{18}=3\sqrt{2}), (\sqrt{50}=5\sqrt{2}), and (\sqrt{8}=2\sqrt{2}), so (y=6\sqrt{2}). Its square is (72).

Multiplying by the conjugate makes the denominator (7-3=4). Hence we get (\frac{2(\sqrt{7}+\sqrt{3})}{4}).

Since ((2+\sqrt{3})(2-\sqrt{3})=1), the reciprocal is (2-\sqrt{3}). Recognizing conjugates is a fast method.

The first product is (25-6=19) and (\sqrt{24}=2\sqrt{6}) is irrational. A rational plus an irrational is irrational.

Rationalize the denominator to find the reciprocal.
\(\dfrac{1}{\sqrt{2}+\sqrt{3}}=\dfrac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^2-(\sqrt{2})^2}=\sqrt{3}-\sqrt{2}\).
Add to \(r\): \(\sqrt{2}+\sqrt{3}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\). Hence the simplified result is \(2\sqrt{3}\).
Why other options are wrong: option C is just \(r\) (not the sum), option D equals \(1/r\), and option B is an incorrect arithmetic result. Exam tip: use conjugates to rationalize denominators whenever square roots appear in the denominator.

Convert each radical to a common √3 factor: \(2\sqrt{12}=2\sqrt{4\cdot3}=4\sqrt{3}\), \(-3\sqrt{27}=-3\sqrt{9\cdot3}=-9\sqrt{3}\), \(\sqrt{75}=\sqrt{25\cdot3}=5\sqrt{3}\). Summing gives \(4\sqrt{3}-9\sqrt{3}+5\sqrt{3}=(4-9+5)\sqrt{3}=0\). Hence the expression simplifies to \(0\). The common distractor \(10\sqrt{3}\) arises from ignoring the negative sign on the middle term, so it is incorrect. Exam tip: factor out perfect squares inside radicals first, then combine like radical terms.

Direct answer: A, rational number. First simplify the radicals: \(\sqrt8=2\sqrt2\) and \(\sqrt{18}=3\sqrt2\). Therefore \(a=5\sqrt2\). Squaring gives \(a^2=(5\sqrt2)^2=25\times2=50\). Since 50 can be written as \(50/1\), it is rational. Option A is correct because rational numbers are numbers expressible as a ratio of integers with a nonzero denominator. Option B is wrong because the irrational parts disappear after squaring. Option C is wrong because square roots of positive numbers are real, and 50 is real. Option D is wrong because 50 is a terminating decimal, namely 50.0, not a non-terminating repeating or non-repeating decimal. The important step is combining like radicals before squaring. Memory cue: simplify first, then square; \(\sqrt8+\sqrt{18}=5\sqrt2\), so the square is 50.

Simplify first: \(\sqrt{45}=3\sqrt{5}\), \(\sqrt{80}=4\sqrt{5}\), \(\sqrt{125}=5\sqrt{5}\). So the expression equals \(3\sqrt{5}+4\sqrt{5}-5\sqrt{5}=2\sqrt{5}\). Since \(\sqrt{5}\) is irrational, multiplying by the nonzero rational 2 yields an irrational number, so \(2\sqrt{5}\) is irrational. Why others are wrong: it cannot be rational because \(\sqrt{5}\) is irrational; it is not an integer; it is not zero because the coefficient 2 is nonzero. Exam tip: factor and combine like surd terms (same \(\sqrt{\;\;}\)) to simplify quickly.

If it were rational, its square (3+\sqrt{5}) would be rational, which it is not. Hence it is real irrational.

Use the identity \((a-b)^2=a^2+b^2-2ab\). With \(a=\sqrt6\) and \(b=\sqrt2\), we get \(a^2+b^2=6+2=8\) and \(2ab=2\sqrt{12}=4\sqrt3\). Thus \(x^2=8-4\sqrt3\). Option B corresponds to a common halving error in the \(2ab\) term; option C has the sign of the middle term wrong. Exam tip: compute \(a^2+b^2\) and \(2ab\) separately to avoid sign or factor mistakes.

The two binomials have the form \((u+v)(u-v)\), which is the difference-of-squares identity \(u^2-v^2\). Using this identity avoids multiplying every term separately and also shows why the square-root terms cancel. The result is rational even though the individual factors contain irrational numbers.

Take \(u=\sqrt{13}\) and \(v=\sqrt{12}\). Then \((\sqrt{13}+\sqrt{12})(\sqrt{13}-\sqrt{12})=(\sqrt{13})^2-(\sqrt{12})^2=13-12=1\). Thus option A is correct. The expression is not 25, and \(\sqrt{156}\) would arise from multiplying only the radicals, not from the complete conjugate product.

Multiplying by the conjugate makes the denominator (5-2=3). So the rationalized form is (\frac{\sqrt{5}-\sqrt{2}}{3}).

(\sqrt{27}=3\sqrt{3}) and (\sqrt{12}=2\sqrt{3}), so the numerator is (5\sqrt{3}). Dividing gives (5).

Actually (x=\sqrt{2}+\sqrt{3}) satisfies (x^4-10x^2+1=0), not a simple quadratic here. Read powers carefully in such trick questions.

The number halfway between two different real numbers is their average. If the numbers are
\(\sqrt{2}\) and \(\sqrt{3}\), their average is \(\frac{\sqrt{2}+\sqrt{3}}{2}\). An average of two numbers lies strictly between them when the numbers are unequal. Therefore option A gives a number between \(\sqrt{2}\) and \(\sqrt{3}\). The other options simplify to 1, 1, and \(\frac12\), respectively, so they are not between these two values.

To verify this numerically, \(\sqrt2\approx1.414\) and \(\sqrt3\approx1.732\). Their average is approximately \(1.573\), which is greater than 1.414 and less than 1.732. Thus the stated answer A is correct. The key idea is that the midpoint of an interval always lies inside that interval, not at either endpoint.

Compute each term separately: \(\sqrt{0.49}=0.7\) and \(\frac{1}{\sqrt{25}}=\frac{1}{5}=0.2\). Their sum is \(0.7+0.2=0.9\). Option B (0.7) represents only \(\sqrt{0.49}\) and omits the second term, so it is incorrect. Exam tip: simplify square roots first (including decimals), then perform the arithmetic to avoid sign or omission errors.

The direct answer is A: \(\sqrt{4n}\) will never be rational under the stated condition. Since 4 is a perfect square, \(\sqrt{4n}=\sqrt4\sqrt n=2\sqrt n\). The given information says \(\sqrt n\) is irrational. Multiplying an irrational number by the nonzero rational number 2 keeps the result irrational: if \(2\sqrt n\) were rational, dividing it by 2 would make \(\sqrt n\) rational, contradicting the premise. Therefore \(\sqrt{4n}\) is irrational for every positive integer n satisfying the condition. Option A is correct. Option B, “always,” is wrong because it claims the result is rational, whereas it is always irrational. Option C is wrong because even when n is even, the given irrationality can remain; for example, n=2 gives \(\sqrt{8}=2\sqrt2\), irrational. Option D is wrong because primality does not change the argument; n=2 is prime and still gives an irrational result. The key exam idea is to factor out the perfect square 4 and observe the nonzero rational multiplier 2.

(\sqrt{8}=2\sqrt{2}), so the sum is (3\sqrt{2}). A non zero rational multiple of (\sqrt{2}) remains irrational.

(\sqrt{98}=7\sqrt{2}) and (\sqrt{18}=3\sqrt{2}). The numerator is (4\sqrt{2}), and division gives (4).

The direct answer is option A: \\(5+2\sqrt6\\). Let the expression be \\(\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}\\). Multiply numerator and denominator by the conjugate of the denominator, \\(\sqrt3+\sqrt2\\). The denominator becomes \\( (\sqrt3)^2-(\sqrt2)^2=3-2=1\\). The numerator becomes \\( (\sqrt3+\sqrt2)^2=3+2+2\sqrt6=5+2\sqrt6\\). Hence the whole value is \\(5+2\sqrt6\\). Option A is correct. Option B, \\(1+\sqrt6\\), misses the two square terms or combines them incorrectly. Option C, \\(5-2\sqrt6\\), would be associated with the square of \\(\sqrt3-\sqrt2\\), not the numerator used here. Option D, \\(\sqrt6\\), omits the rational terms. Memory cue: rationalise a radical denominator with its conjugate; then expand carefully.

Multiplying by the conjugate gives denominator (5-3=2) and numerator (8+2\sqrt{15}). The simplified form is (4+\sqrt{15}).

Write each radical in simplest form: \sqrt{12}=2\sqrt{3}, \sqrt{27}=3\sqrt{3}, \sqrt{75}=5\sqrt{3}, \sqrt{48}=4\sqrt{3}. Adding and subtracting gives 2\sqrt{3}+3\sqrt{3}+5\sqrt{3}-4\sqrt{3}=6\sqrt{3}, so A is correct. Options B and C reflect common arithmetic mistakes in combining coefficients; option D (\sqrt{66}) is not possible here because you must combine like radicals (same radicand) after simplifying. Exam tip: factor out perfect square factors first, then combine like radical terms by adding their coefficients.

The decimal expansion is infinite and the gaps of zeros between successive 1s increase: patterns of 1 appear at positions 1, 3, 6, 10, ... (triangular numbers). A rational number must have a decimal expansion that eventually becomes periodic (a fixed repeating block). Since no such periodic block exists here, the number is irrational. The closest distractor, rational number, is incorrect because there is no eventual repetition; terminating decimal is wrong because the expansion does not end; integer is impossible because the value lies between 0 and 1. Exam tip: to test for irrationality, look for eventual periodicity — if none exists and the expansion is infinite, the number is irrational.