Irrational numbers and real numbers

The governing concept is the use of conjugate surds and the difference-of-squares identity. Add the two expressions directly: (3+√8)+(3−√8)=3+3+√8−√8=6, because the radical terms cancel. For the product, use (a+b)(a−b)=a²−b². Thus (3+√8)(3−√8)=3²−(√8)²=9−8=1. Therefore option A gives both required values. Option B incorrectly treats the two expressions as opposites, while option C does not cancel the conjugate terms correctly. Option D has the correct sum but wrongly gives the product as 17. The irrationality of √8 does not prevent the product of conjugates from being rational.

Use the square formula \((a-b)^2=a^2+b^2-2ab\). With \(a=\sqrt{7}\) and \(b=\sqrt{3}\) we get \((\sqrt{7}-\sqrt{3})^2=7+3-2\sqrt{7}\sqrt{3}=10-2\sqrt{21}\), so option A is correct. The closest wrong choice \(10+2\sqrt{21}\) has the wrong sign for the \(2ab\) term. Option C \(4-2\sqrt{21}\) arises from confusing signs or subtracting instead of adding \(a^2\) and \(b^2\). Exam tip: always square each term first, then compute the \(-2ab\) term and check its sign before simplifying radicals.

A terminating decimal can be expressed as an integer divided by a power of 10, e.g. as \(\frac{p}{10^n}\). After cancelling common factors this is always of the form \(\frac{a}{b}\) with integers a,b, which is the definition of a rational number. Hence every terminating decimal is rational. Closest distractor: "always an integer" is incorrect because examples like 0.5 or 1.25 are terminating decimals but not integers. Exam tip: convert a terminating decimal to \(\frac{p}{10^n}\) and simplify to quickly show it's rational (or to compare values).

Rational numbers can be expressed as a fraction p/q (integers p and q, q ≠ 0). The decimal expansion of any rational number is either terminating or non‑terminating repeating. Examples: 0.333... = 1/3 and 0.2727... = 27/99 = 3/11. Thus a non‑terminating repeating decimal is rational. Why other options are wrong: B (irrational) is incorrect because irrational numbers have non‑terminating, non‑repeating decimals (e.g. √2). C (non‑real) is incorrect since such decimals represent real numbers. D (integer only) is incorrect because repeating decimals are generally fractions, not necessarily integers. Exam tip: Convert the repeating decimal into a fraction using algebra (e.g., set x = 0.2727..., multiply to shift decimal, subtract) to verify rationality quickly.

\((\frac{2}{5})\) is rational (a ratio of integers) while \(\sqrt{17}\) is irrational because 17 is not a perfect square. If the sum were rational, then \(\sqrt{17}=(\frac{2}{5}+\sqrt{17})-\frac{2}{5}\) would be rational too, which is impossible. Hence the sum is irrational. The closest distractor is "Rational": this would require the irrational part to cancel out, which does not happen here. Options "Integer" and "Terminating decimal" are special cases of rational numbers and are therefore also impossible. Exam tip: isolate the irrational term — subtract any rational parts; if an irrational like \(\sqrt{m}\) with m not a perfect square remains, the whole expression is irrational.

Note that \(\sqrt{275}=\sqrt{25\times11}=5\sqrt{11}\). Therefore \(5\sqrt{11}-\sqrt{275}=5\sqrt{11}-5\sqrt{11}=0\). Option B (\(10\sqrt{11}\)) is incorrect because it is twice the term, option C (\(5\sqrt{11}\)) is just the first term without subtraction, and option D (\(\sqrt{11}\)) is only one-fifth of the term. Exam tip: always simplify radicals by factoring out perfect squares before performing addition or subtraction of surds.

Factor the radicand: \(\sqrt{432}=\sqrt{144\times3}\). Since \(144=12^2\), extract the square root: \(\sqrt{432}=\sqrt{144}\,\sqrt{3}=12\sqrt{3}\). Options B and C are not equal to \(12\sqrt{3}\) numerically; option D simplifies to \(3\sqrt{144}=3\times12=36\), which is incorrect. Exam tip: always look for the largest perfect square factor of the number under the radical and pull its square root outside first.

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

This is a difference-of-squares: let a=\sqrt{10}, b=\sqrt{5}. Then (a+b)(a-b)=a^2-b^2. Here a^2=10 and b^2=5, so the value is 10-5=5. Option C (\sqrt{50}) is incorrect because \sqrt{50}=5\sqrt{2}, not 5; option B (1) and D (15) are also wrong — 1 could come from an erroneous division and 15 from adding 10 and 5. Exam tip: Recognise and apply (a+b)(a-b)=a^2-b^2 to simplify quickly without expanding radicals.

Use the identity \((a+b)^2=a^2+2ab+b^2\). Here \(a=2\sqrt{3}\) and \(b=1\). Compute: \((2\sqrt{3})^2=4\cdot3=12,\; 2ab=2\cdot(2\sqrt{3})\cdot1=4\sqrt{3},\; b^2=1\). Summing gives \(12+4\sqrt{3}+1=13+4\sqrt{3}\). Option B (\(12+4\sqrt{3}\)) is the closest wrong choice — it omits the \(+1\). Options C and D have incorrect coefficients for the \(\sqrt{3}\) term or the middle term. Exam tip: Always apply the formula first, simplify each term separately, then add.

Compute square factors: \(\sqrt{27}=\sqrt{9\cdot3}=3\sqrt{3}\) and \(\sqrt{75}=\sqrt{25\cdot3}=5\sqrt{3}\). So \(\sqrt{3}+\sqrt{27}+\sqrt{75}=\sqrt{3}+3\sqrt{3}+5\sqrt{3}=(1+3+5)\sqrt{3}=9\sqrt{3}\). Choice B (\(\sqrt{105}\)) is a common mistake of combining radicals under one root; you can only do that when appropriate (e.g., same radicand). Choices C and D result from incorrect addition of coefficients. Exam tip: always factor out perfect squares from inside radicals first, then combine like radical terms by adding their coefficients.

\(\sqrt{363}=\sqrt{121\times3}=\sqrt{121}\times\sqrt{3}=11\sqrt{3}\). Thus the simplified form is \(11\sqrt{3}\). The closest distractor \(3\sqrt{11}\) is incorrect because \(3\sqrt{11}=\sqrt{9\times11}=\sqrt{99}\), not \(\sqrt{363}\). Options \(\sqrt{33}\) and \(\dfrac{\sqrt{121}}{\sqrt{3}}\) evaluate to different values (the last equals \(11/\sqrt{3}\)). Exam tip: always factor the radicand and pull out the largest perfect square factor to simplify roots quickly.

(\sqrt{243}=9\sqrt{3}), (\sqrt{147}=7\sqrt{3}), and (\sqrt{75}=5\sqrt{3}). The result is (11\sqrt{3}).

The governing concept is comparing square roots by comparing their non-negative radicands, followed by checking whether the result is rational or irrational. Since 10=√100 and 11=√121, and 100<115<121, taking positive square roots gives 10<√115<11. Also, 115 is not a perfect square, so √115 cannot be written as a ratio of integers and is irrational. Thus option A satisfies both conditions. Option B equals √100=10, so it is not strictly between the endpoints. Option C equals √121=11 and is also an endpoint. Option D equals 10.5, which is between 10 and 11 but is rational, not irrational. Therefore only A is valid.

A rational number has a decimal expansion that is eventually periodic (repeating). In 9.0202202220... the zeros occur at positions 1, 3, 6, 10, ... which are the triangular numbers \\(T_n=\frac{n(n+1)}{2}\\). The blocks of consecutive 2's between zeros have lengths 1, 2, 3, 4, ... which grow without bound, so no fixed repeating block exists. Hence the decimal is non-terminating and non-repeating, and the number is irrational. Exam tip: check whether gaps between recognizable markers (here zeros) stabilize; if they keep changing indefinitely, the decimal cannot be periodic and the number is irrational.

The direct answer is option A: 38. The expressions p and q are conjugates: \\(p=7+\sqrt{11}\\) and \\(q=7-\sqrt{11}\\). Their product uses \\( (a+b)(a-b)=a^2-b^2\\). Thus \\(pq=7^2-(\sqrt{11})^2=49-11=38\\). Option A is correct. Option B, 60, would result from incorrectly adding 49 and 11. Option C, \\(14\sqrt{11}\\), is not the product because the two cross terms cancel: \\(7\sqrt{11}-7\sqrt{11}=0\\). Option D, \\(49+\sqrt{11}\\), fails to square the radical and does not apply the product formula. The irrational parts disappear, leaving a rational number. Memory cue: conjugate products remove the middle terms and become “square minus square.”

The governing idea is expansion of a binomial containing surds. Substitute x = \sqrt{5}+\sqrt{2} and square it: x^2 = (\sqrt{5}+\sqrt{2})^2 = 5+2+2\sqrt{5}\sqrt{2} = 7+2\sqrt{10}. Thus x^2-2\sqrt{10} = (7+2\sqrt{10})-2\sqrt{10}=7. Since 7 is an integer, it is also a rational real number. Option B is incorrect because the irrational surd terms cancel exactly. Option C is impossible because the expression is formed from real numbers, and option D does not describe 7. Therefore option A is the unique correct answer.

The governing concept is rationalizing a denominator containing a surd. The conjugate of 3+√5 is 3−√5, so multiply both numerator and denominator by this conjugate: 1/(3+√5)=(3−√5)/[(3+√5)(3−√5)]. Applying the difference-of-squares identity makes the denominator 3²−(√5)²=9−5=4. Therefore the rationalized form is (3−√5)/4, which is option A. Option B uses the original denominator expression rather than its conjugate, so it does not produce the required cancellation. Option C has an incorrect denominator, and option D omits the irrational part of the numerator and is not equal to the original fraction. Thus A is the unique correct answer.

The governing concept is forming a polynomial equation satisfied by a surd using its conjugate. If x=2+√3, its conjugate is 2−√3. The sum of these two numbers is 4, and their product is (2+√3)(2−√3)=4−3=1. A monic quadratic whose roots have sum 4 and product 1 is t²−4t+1=0, because the general form is t²−(sum of roots)t+(product of roots)=0. Since x is one of the roots, x²−4x+1=0. Hence A is correct. Option B would describe a repeated root 1, option C has the wrong middle-term sign, and option D has an incorrect root sum and product.

Answer: A, 8. A rational number can be written as p/q, where p and q are integers and q is not zero; every integer is therefore rational. To check the inequality, compare squares because all the numbers here are positive. We know 7² = 49, 8² = 64, and 9² = 81. Since 49 < 50 < 64, we get 7 < √50 < 8. Since 64 < 72 < 81, we get 8 < √72 < 9. Thus √50 < 8 < √72, so A satisfies the inequality and is rational. B, √60, lies between the bounds, but 60 is not a perfect square, so √60 is irrational. C, 7, is below √50 because 49 < 50. D, 9, is above √72 because 81 > 72. Memory cue: square positive quantities to compare roots, but separately check whether the chosen number is rational.

A zero of a polynomial is a value of x for which the polynomial becomes zero. Set 5x²−5=0. Factoring gives 5(x²−1)=0, so x²−1=0 and therefore (x−1)(x+1)=0. Hence x=1 or x=−1. Both numbers are integers, and every integer is rational because it can be written as a quotient of two integers, such as 1=1/1 and −1=−1/1. The common factor 5 does not produce √5; it is removed by division. Thus option A is correct. Option C confuses coefficients with roots, option B results from an incorrect rearrangement, and option D is false because two real roots exist.

Use Vieta’s relations for a quadratic ax²+bx+c. If its zeroes are α and β, then α+β=−b/a and αβ=c/a. In p(x)=x²−2√6x+5, the coefficients are a=1, b=−2√6, and c=5. Hence the sum is −(−2√6)/1=2√6, while the product is 5/1=5. Thus option A gives both values in the requested order. Option B interchanges the sum and product. Option C incorrectly changes the sign of the product even though the constant term is positive. Option D forgets the negative sign in the formula for the sum. Individual roots are unnecessary because the coefficient relations answer the question directly.

For a monic quadratic x²+bx+c, the product of its zeroes equals the constant term c. The given zeroes are 1+√6 and 1−√6, so use the difference-of-squares identity: (1+√6)(1−√6)=1²−(√6)²=1−6=−5. Therefore m, the constant term, is −5. The sum provides an additional check: (1+√6)+(1−√6)=2, and the coefficient of x is −2, exactly the negative of that sum. Thus option A is correct. Option B loses the negative sign. The values 7 and −7 may result from incorrectly adding the radical terms or mishandling the product; neither agrees with the coefficient relations. Both Vieta’s product and sum confirm the answer.

For a monic quadratic x²+px−3 with zeroes α and β, Vieta’s relations give α+β=−p and αβ=−3. Add the two given zeroes: (2+√7)+(2−√7)=4, because the irrational terms cancel. Therefore −p=4, so p=−4. The product independently checks the information: (2+√7)(2−√7)=2²−(√7)²=4−7=−3, which agrees with the constant term. Hence option A is correct. Option B wrongly treats the sum as p rather than −p. Options C and D confuse p with the constant term −3 or mishandle its sign. Both root relations confirm the same answer.

For a quadratic ax²+bx+c with zeroes α and β, Vieta’s relation gives αβ=c/a. In p(x)=x²−2x−3√2, a=1 and c=−3√2. Therefore αβ=(−3√2)/1=−3√2, so option A is correct. The sum follows a different relation, α+β=−b/a; here it equals 2, not −3√2. The discriminant is b²−4ac, which becomes (−2)²−4(1)(−3√2)=4+12√2, so option D is also incorrect. Finally, an irrational constant term by itself does not prove that both zeroes are rational or irrational; more information would be needed. The key point is that the monic constant term equals the product of the zeroes.