Irrational numbers

Correct answer: A. (\frac{\sqrt{13}-3}{4})

Explanation: Multiplying by the conjugate (\sqrt{13}-3) makes the denominator (13-9=4). So (p=\frac{\sqrt{13}-3}{4}).

Correct answer: A. (9+\sqrt{20})

Explanation: The direct answer is A, 9+√20. Let x = √(9+√20). The expression is x×x, and the basic rule is √a×√a = a whenever a is non-negative. Here 9+√20 is positive, so the product is exactly 9+√20. Option A is correct. Option B, 81+20 = 101, incorrectly squares the two parts separately and ignores the middle term in a square. Option C, √29, is not equal to 9+√20; it incorrectly combines the numbers under different radicals. Option D, 9−√20, is the conjugate-like expression and is not produced by multiplying the same square root by itself. No extra expansion is needed. A useful check is to call the whole quantity inside the radical one number rather than splitting it.

Correct answer: C. It can be either rational or irrational.

Explanation: Option C is correct. For \(a=\sqrt{2}, b=-\sqrt{2}\), the sum is 0 and \(ab=-2\) is rational. But with \(b=1-\sqrt{2}\), the sum is 1 while \(ab=\sqrt{2}-2\) is irrational. Exam tip: verify a claim using contrasting examples.

Correct answer: A. (9+4\sqrt{5})

Explanation: The direct answer is A: \(9+4\sqrt5\). Rationalise the denominator by multiplying by its conjugate, \(\sqrt5+2\): \(\frac{\sqrt5+2}{\sqrt5-2}\times\frac{\sqrt5+2}{\sqrt5+2}=\frac{(\sqrt5+2)^2}{5-4}\). The denominator is 1, so the result is \((\sqrt5+2)^2\). Expanding gives \(5+4\sqrt5+4=9+4\sqrt5\). Option A is correct. Option B, \(9-4\sqrt5\), results from using the wrong sign in the expansion and is also the value of a different squared expression. Option C, \(1+\sqrt5\), does not follow from the square. Option D, \(5+2\sqrt5\), omits the final constant 4 and uses an incorrect middle coefficient. The safe method is conjugate first, then expand carefully.

Correct answer: A. (\frac{11+4\sqrt{6}}{5})

Explanation: Multiplying by the denominator conjugate gives denominator (8-3=5). The numerator is ((\sqrt{8}+\sqrt{3})^2=11+4\sqrt{6}).

Correct answer: A. \(a+b\) is irrational

Explanation: Assume \(a+b\) is rational. Then \(a=(a+b)-b\) would be the difference of two rational numbers and hence rational, a contradiction. Therefore \(a+b\) is irrational. Exam tip: rational ± irrational is always irrational.

Correct answer: B. Its square is (10+\sqrt{21})

Explanation: Since (10+\sqrt{21}) is positive, its square root is real. Squaring it gives the inside number.

Correct answer: A. \(x+r\) is irrational

Explanation: If \(x+r\) were rational, then \(x=(x+r)-r\) would be the difference of two rational numbers and hence rational, a contradiction. Thus A is correct. Exam tip: rational ± irrational is always irrational.

Correct answer: A. (\frac{2(\sqrt{15}+\sqrt{6})}{3})

Explanation: Multiplying by the conjugate makes the denominator (15-6=9). So (\frac{6(\sqrt{15}+\sqrt{6})}{9}=\frac{2(\sqrt{15}+\sqrt{6})}{3}).

Correct answer: A. (288)

Explanation: The direct answer is option A, 288. Simplify each radical first: √50 = √(25 × 2) = 5√2, and √98 = √(49 × 2) = 7√2. Therefore √50 + √98 = 5√2 + 7√2 = 12√2. Now square the sum: (12√2)² = 12² × (√2)² = 144 × 2 = 288. Option A, 288, is correct. Option B, 148, is wrong because it does not result from correctly squaring 12√2. Option C, 24√2, is wrong because it is not the value of the square; it resembles an unsquared or incorrectly expanded result. Option D, 196, is wrong because it could arise from squaring 14, but the simplified sum is 12√2, not 14. One may also check by expansion: 50 + 98 + 2√(50×98) = 148 + 2√4900 = 148 + 140 = 288. The useful cue is to extract perfect-square factors before adding like radicals.

Correct answer: B. एक संख्या दूसरी संख्या का योगात्मक प्रतिलोम हो

Explanation: If one irrational number is x and the other is its additive inverse −x, then \(x+(-x)=0\). Since 0 is rational, the sum is rational. Exam tip: for “guaranteed” statements, check whether every allowed case works, not just one example.

Correct answer: A. (10)

Explanation: \(\sqrt{125}=5\sqrt{5}\) and \(\sqrt{45}=3\sqrt{5}\), so the bracket is \(2\sqrt{5}\). Multiplying by \(\sqrt{5}\) gives (10).

Correct answer: A. (\sqrt{12}-\sqrt{7})

Explanation: Multiplying by the conjugate makes the denominator (12-7=5). So (\frac{5(\sqrt{12}-\sqrt{7})}{5}=\sqrt{12}-\sqrt{7}).

Correct answer: A. The product of a non-zero rational number and an irrational number is irrational.

Explanation: If \(q\ne0\) is rational and \(x\) is irrational, assuming \(qx\) rational gives \(x=(qx)/q\) rational, a contradiction. B fails because \(\sqrt{2}+(-\sqrt{2})=0\). Exam tip: test “always” claims with a counterexample.

Correct answer: A. (13\sqrt{2})

Explanation: (\sqrt{242}=11\sqrt{2}), (\sqrt{128}=8\sqrt{2}), and (\sqrt{72}=6\sqrt{2}). Therefore the result is (13\sqrt{2}).

Correct answer: A. (2\sqrt{13})

Explanation: The direct answer is option A, 2√13. Given t = √13 + 3, find 4/t by rationalising the denominator: 4/(√13 + 3) × (√13 − 3)/(√13 − 3). The denominator is 13 − 9 = 4, so the fraction becomes 4(√13 − 3)/4 = √13 − 3. Hence t + 4/t = (√13 + 3) + (√13 − 3) = 2√13. Option A is correct because the rationalised fraction cancels the +3 and −3 terms. Option B, 6, is wrong because the radical terms do not cancel; they add to 2√13. Option C, 2√13 + 6, is wrong because it incorrectly adds the constants instead of noticing that +3 and −3 cancel. Option D, √13, is wrong because there are two √13 terms, not one. The key idea is that √13 + 3 and √13 − 3 are conjugates, and their product is 13 − 9 = 4. This makes the fraction easy to simplify.

Correct answer: A. (13)

Explanation: (\sqrt{75}=5\sqrt{3}) and (\sqrt{192}=8\sqrt{3}), so (s=13\sqrt{3}). Dividing by (\sqrt{3}) gives (13).

Correct answer: A. (\frac{5-\sqrt{21}}{2})

Explanation: Multiplying by the conjugate gives (\frac{(\sqrt{7}-\sqrt{3})^2}{7-3}). So the answer is (\frac{5-\sqrt{21}}{2}).

Correct answer: C. If \(r\) is a non-zero rational number and \(x\) is irrational, then \(r+x\) is irrational.

Explanation: Option C is correct. If \(r+x\) were rational, then \(x=(r+x)-r\) would also be rational, a contradiction. For A, \(\sqrt{2}+(-\sqrt{2})=0\). Exam tip: test “always” claims using a counterexample.

Correct answer: A. \(\sqrt{2},\ \sqrt{3},\ \sqrt{5}\)

Explanation: Since 2, 3 and 5 are not perfect squares, \(\sqrt2\), \(\sqrt3\) and \(\sqrt5\) are all irrational. The other options contain \(\sqrt{49}=7\), \(\sqrt2\times\sqrt8=\sqrt{16}=4\), or \(\frac{\sqrt{12}}{\sqrt3}=\sqrt4=2\), which are rational. Exam tip: a square root of a perfect square is rational.

Correct answer: A. (14)

Explanation: Area is ((5+\sqrt{11})(5-\sqrt{11})=25-11=14). Multiplying conjugate dimensions can give a rational area.

Correct answer: A. (12\sqrt{5})

Explanation: (\sqrt{500}=10\sqrt{5}), (\sqrt{320}=8\sqrt{5}), and (\sqrt{180}=6\sqrt{5}). So the result is (10\sqrt{5}-8\sqrt{5}+6\sqrt{5}=8\sqrt{5}).

Correct answer: A. (2\sqrt{8})

Explanation: The first term becomes (\frac{\sqrt{8}-\sqrt{6}}{2}) and the second becomes (\frac{\sqrt{8}+\sqrt{6}}{2}). Their sum is (\sqrt{8}).

Correct answer: B. \(\sqrt{5}+(3-\sqrt{5})\)

Explanation: Both \(\sqrt{5}\) and \(3-\sqrt{5}\) are irrational, but their sum is \(\sqrt{5}+3-\sqrt{5}=3\), which is rational. Thus the claim is false. Exam tip: check whether irrational terms cancel.

Correct answer: A. \(\sqrt{7}+(-\sqrt{7})=0\)

Explanation: Both \(\sqrt{7}\) and \(-\sqrt{7}\) are irrational, but they are additive inverses, so their sum is \(0\), a rational number. Thus the word “always” makes the claim false. Exam tip: one counterexample is enough to disprove a universal statement.