Irrational numbers

(\sqrt{726}=11\sqrt{6}), (\sqrt{486}=9\sqrt{6}), and (\sqrt{150}=5\sqrt{6}). Therefore the result is (7\sqrt{6}).

To rationalise a denominator containing a difference of square roots, multiply the fraction by the conjugate of that denominator. The conjugate of \(\sqrt{43}-\sqrt{31}\) is \(\sqrt{43}+\sqrt{31}\). Thus the denominator becomes \((\sqrt{43})^2-(\sqrt{31})^2=43-31=12\), which is rational.

The numerator then becomes \(12(\sqrt{43}+\sqrt{31})\). Cancelling the common factor 12 with the denominator leaves \(\sqrt{43}+\sqrt{31}\). Hence option C is correct. Option A reverses the factor instead of multiplying by it, while option D uses an incorrect product of the radicands. The supplied answer and explanation are correct.

If rx were rational, dividing it by the non-zero rational number r would make x rational, a contradiction. Hence rx is irrational. Option A fails because √2 + (−√2) = 0. Exam tip: test “always” statements using a counterexample.

Area is ((\sqrt{41}+\sqrt{17})(\sqrt{41}-\sqrt{17})=41-17=24). Conjugate dimensions give rational area.

(\frac{\sqrt{75}}{\sqrt{3}}=\sqrt{25}=5), which is rational. Check whether division forms a perfect square.

Both \(\sqrt{2}\) and \(-\sqrt{2}\) are irrational, but they cancel to give \(0\), which is rational. Hence “always” is false. Exam tip: one counterexample is enough to disprove a universal statement.

Since \(3\) is not a perfect square, \(\sqrt{3}\) is irrational. The decimal expansion of an irrational number is non-terminating and non-recurring, so option C is correct. Option B is wrong because a non-terminating recurring decimal represents a rational number, such as \(0.333\ldots\). Exam tip: The square root of a positive integer is rational only if the integer is a perfect square.

Let \(x+y=q\), where \(q\) is rational. Then \(y=q-x\). A rational number minus an irrational number is irrational, so B is necessary. Exam tip: rearrange the equation before classifying numbers.

Both \(\sqrt{2}\) and \(-\sqrt{2}\) are irrational, but their sum is \(0\), which is rational. In the other pairs, an irrational radical remains. Exam tip: first look for additive inverses.

Let \(r\) be rational and \(x\) be irrational. Then \(r+x\) must be irrational. If \(r+x\) were rational, then \(x=(r+x)-r\) would be the difference of two rational numbers and hence rational, which is a contradiction. For example, \(3+\sqrt{2}\) is irrational. Therefore, option A is incorrect, and the sum is not necessarily an integer either. Exam tip: the sum or difference of a rational number and an irrational number is always irrational.

\(\sqrt{8}=\sqrt{4\times2}=2\sqrt{2}\). Therefore, \(\sqrt{2}+\sqrt{8}=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\). Option B results from incorrectly combining the numbers inside different radicals, while options C and D give an incorrect value or only \(\sqrt{8}\). Exam tip: square-root terms can be added or subtracted like algebraic terms only when their radicands are the same after simplification.

Since \(3^2=9<10<16=4^2\), we have \(3<\sqrt{10}<4\). Also, \(4^2=16<17<25=5^2\), so \(4<\sqrt{17}<5\). Hence, \(4\) is greater than \(\sqrt{10}\) and less than \(\sqrt{17}\). Option \(3\) is less than \(\sqrt{10}\), while \(5\) and \(6\) are greater than \(\sqrt{17}\). Exam tip: locate a number between consecutive perfect squares to compare its square root.

The number of zeros between successive 1s keeps increasing, so no fixed digit block repeats. Its decimal is non-terminating and non-recurring; hence it is irrational. Exam tip: check for a repeating block.

Since \(\sqrt{75}=\sqrt{25\times3}=5\sqrt{3}\), we get \(\sqrt{75}\div\sqrt{3}=5\sqrt{3}\div\sqrt{3}=5\). Option B incorrectly treats the radicand 25 as the final value, while option C results from an incorrect subtraction of radicands. Exam tip: factor the radicand into a perfect square and another factor before simplifying.

\(2\) is rational, while \(\sqrt{3}\) is irrational because 3 is not a perfect square. The sum of a rational number and an irrational number is always irrational. Therefore, \(x=2+\sqrt{3}\) is irrational. Integers and natural numbers are rational, so neither can be correct here. Exam tip: rational \(+\) irrational is always irrational.

Since \(27=9\times3\) and \(12=4\times3\), \(\sqrt{27}=3\sqrt{3}\) and \(\sqrt{12}=2\sqrt{3}\). Thus, \(3\sqrt{3}+2\sqrt{3}-\sqrt{3}=4\sqrt{3}\). The distractor \(5\sqrt{3}\) results from forgetting to subtract the final \(\sqrt{3}\). Exam tip: extract perfect-square factors first, then combine like surds by adding or subtracting their coefficients.

Here, \(5\) is a non-zero rational number and \(\sqrt{x}\) is irrational. The product of a non-zero rational number and an irrational number is always irrational. Therefore, \(5\sqrt{x}\) is an irrational number. An integer and zero are both rational, so they cannot be the result here. Exam tip: Multiplication by zero is the exception that can make the product rational.

For nonsquare \(n\), \(\sqrt n\) is irrational. If \(a+\sqrt n\) were rational, subtracting rational \(a\) would make \(\sqrt n\) rational—a contradiction. Tip: \(\sqrt{a^2}=|a|\) is rational.

\(\sqrt{32}=\sqrt{16\times2}=4\sqrt{2}\) and \(\sqrt{18}=\sqrt{9\times2}=3\sqrt{2}\). Therefore, \(\sqrt{32}+\sqrt{18}=4\sqrt{2}+3\sqrt{2}=7\sqrt{2}\), so option B is correct. Exam tip: Factor out the largest perfect-square factor before adding surds.

For non-negative \(a\) and \(b\), squaring the two sides gives \(a+b+2\sqrt{ab}\) on the left and \(a+b\) on the right. They are equal only when \(2\sqrt{ab}=0\), that is, when \(ab=0\). Hence, this is not a general rule; it works only in special cases. Option D describes one such special case, so it cannot represent the general conclusion. For example, \(\sqrt{4}+\sqrt{9}=5\), whereas \(\sqrt{4+9}=\sqrt{13}\). Exam tip: do not directly combine a sum of square roots into one square root.

Since \(9<11<15<16\), we have \(3<\sqrt{11}<\sqrt{15}<4\). Both square roots lie between 3 and 4, and there is no other whole number between these consecutive whole numbers. Thus, 3 is smaller than \(\sqrt{11}\), while 4 is greater than \(\sqrt{15}\). Exam tip: compare with nearby perfect squares to locate square roots quickly.

\(\sqrt{121}=11\), which is rational, while \(\sqrt{2}\) is irrational. The sum of a rational number and an irrational number is always irrational. Hence, \(\sqrt{121}+\sqrt{2}=11+\sqrt{2}\) is irrational. Integers and natural numbers are rational, so options C and D cannot be correct. Exam tip: the square root of a perfect square is an integer, but \(2\) is not a perfect square.