Irrational numbers and real numbers

Correct answer: A. \(5\sqrt{6}\)

Explanation: Correct because \(\sqrt{150}=\sqrt{25\times6}=5\sqrt{6}\); 25 is the largest perfect square factor and can be taken outside the root. Why the closest distractor is wrong: \(3\sqrt{50}=3\times5\sqrt{2}=15\sqrt{2}\), not equal to \(5\sqrt{6}\). Exam tip: always factor the radicand into the largest perfect square times the remainder to simplify square roots quickly.

Correct answer: A. (10\sqrt{2})

Explanation: (\sqrt{200}=\sqrt{100\times2}=10\sqrt{2}). Taking out the greatest perfect square is a good method.

Correct answer: A. \(4\sqrt{2}\)

Explanation: Since \(\sqrt{18}=\sqrt{9\times2}=3\sqrt{2}\), we have \(\sqrt{2}+\sqrt{18}=\sqrt{2}+3\sqrt{2}=4\sqrt{2}\). Option B, \(\sqrt{20}=2\sqrt{5}\), is not equal to \(4\sqrt{2}\). Option C equals only \(\sqrt{18}\) (i.e. \(3\sqrt{2}\)), and option D is too small. Exam tip: always simplify each radical by extracting perfect squares first, then combine like (same-radical) terms.

Correct answer: A. \(6\sqrt{3}\)

Explanation: Since \(\sqrt{75}=\sqrt{25\times3}=5\sqrt{3}\), we have \(\sqrt{3}+\sqrt{75}=\sqrt{3}+5\sqrt{3}=6\sqrt{3}\). The common wrong idea \(\sqrt{78}\) assumes \(\sqrt{a}+\sqrt{b}=\sqrt{a+b}\), which is not true in general. Exam tip: factor out perfect squares first and then add like radical terms.

Correct answer: A. It is rational and real

Explanation: Zero can be expressed as \\(0=\frac{0}{1}\\), so it is the ratio of two integers and therefore rational. Zero also lies on the number line, so it is a real number. Option B is incorrect because irrational numbers cannot be written as a ratio of integers, but zero can. Option C is incorrect because zero is included among real numbers. Option D is misleading: many conventions define natural numbers as 1,2,3,... so zero is not necessarily a natural number (but check the definition used). Exam tip: always confirm the convention for 'natural numbers' in the question context or syllabus.

Correct answer: A. It is non-terminating and non-repeating

Explanation: The decimal is formed by concatenating the natural numbers: 0.1 2 3 4 5 6 7 8 9 10 11 12 ... . It never terminates and does not settle into any fixed periodic block, so it is non‑terminating and non‑repeating. A non‑terminating, non‑repeating decimal represents an irrational number. The closest distractor (C) is wrong because a repeating decimal would show a constant repeating pattern after some point, which this sequence does not. Options B and D are also incorrect: it does not terminate and is not an integer. Exam tip: to test rationality, look for a repeating block — its presence implies rationality, its absence (for infinite decimals) implies irrationality.

Correct answer: A. ({2,-5,0.4,\frac{7}{8}})

Explanation: All numbers in the first set can be written in (\frac{p}{q}) form. The other sets contain an irrational number.

Correct answer: A. ({\sqrt{3},\sqrt{5},\sqrt{12}})

Explanation: (\sqrt{3}), (\sqrt{5}), and (\sqrt{12}) are all irrational. Remove options with perfect squares and rational numbers.

Correct answer: A. Rational number

Explanation: A number of the form \(\frac{p}{q}\) with integers \(p,q\) and \(q\neq0\) is, by definition, a rational number. Rational numbers have decimal expansions that either terminate or repeat. Option C (real number) is true in a broader sense because every rational is real, but it is not the specific name asked for. Option B (irrational) is incorrect since irrationals cannot be written as \(\frac{p}{q}\). Option D (natural number) is incorrect because natural numbers are positive integers, not general fractions. Exam tip: remember the defining condition — expressible as \(\frac{p}{q}\) with integer numerator and nonzero integer denominator; look for terminating/repeating decimals as a quick check.

Correct answer: A. Irrational number

Explanation: The direct answer is option A: irrational number. Real numbers include both rational and irrational numbers. A rational number can be written as \\(p/q\\), where p and q are integers and \\(q\\neq0\\). A real number that cannot be written in this form is called irrational. Therefore the description in the question directly defines an irrational number. Option A is correct. Option B, integer, is too narrow and is actually always rational, since any integer m can be written as \\(m/1\\). Option C, natural number, is also a special kind of integer and therefore rational. Option D, whole number, is another nonnegative-integer category and is also rational. Examples of irrational real numbers are \\(\sqrt2\\), \\(\pi\\), and \\(\sqrt5\\); their decimals do not terminate or repeat in a fixed cycle. Memory cue: real numbers split into rational and irrational; “not rational” means irrational.

Correct answer: A. It is an irrational number

Explanation: Since 11 is not a perfect square, \(\sqrt{11}\) is irrational. The difference of a rational number (3) and an irrational number is irrational: if \(3-\sqrt{11}\) were rational then \(\sqrt{11}=3-(3-\sqrt{11})\) would be rational, a contradiction. Numerically \(3-\sqrt{11}\approx-0.316\), so it is neither zero nor an integer. Exam tip: check whether the radical term cancels exactly; if it doesn't, the expression remains irrational.

Correct answer: A. (\sqrt{256}) is rational and (\sqrt{257}) is irrational

Explanation: The direct answer is A: \(\sqrt{256}\) is rational and \(\sqrt{257}\) is irrational. A rational number can be written as a fraction of integers; every integer is rational. Since \(256=16^2\), \(\sqrt{256}=16\), which is an integer and therefore rational. For 257, the nearby squares are \(16^2=256\) and \(17^2=289\). Because 257 lies strictly between these consecutive perfect squares, it is not a perfect square. The square root of a positive integer is rational only when the integer is a perfect square, so \(\sqrt{257}\) is irrational. Option A gives exactly this pair. Option B is wrong because 257 is not a perfect square. Option C is wrong because \(\sqrt{256}=16\) is rational. Option D reverses both classifications and is therefore wrong. Remember: first check whether the number under a square root is a perfect square; if yes, the root is rational, and if not, it is irrational.

Correct answer: A. \(11\sqrt{2}\)

Explanation: \(\sqrt{242}=\sqrt{121\times2}=\sqrt{121}\times\sqrt{2}=11\sqrt{2}\). Thus the simplified form is \(11\sqrt{2}\). Option B is a common mistake: \(2\sqrt{11}=\sqrt{4\times11}=\sqrt{44}\), not \(\sqrt{242}\). Option C wrongly splits a sum under a root (\(\sqrt{a+b}\neq\sqrt{a}+\sqrt{b}\)). Option D has an incorrect factor. Exam tip: always factor out the largest perfect square first (here 121).

Correct answer: A. Rational number

Explanation: The surd terms cancel: \((5+\sqrt{7})-(2+\sqrt{7})=5-2+(\sqrt{7}-\sqrt{7})=3\). The result is an integer, hence a rational number (\(3=3/1\)). Option B is incorrect because an irrational number cannot be written as a ratio of integers; here the result is an integer. Option C is incorrect because non‑real numbers have imaginary parts, which are absent. Option D is incorrect because the result is positive. Exam tip: combine like terms and cancel identical surds first — it quickly simplifies such expressions.

Correct answer: A. \(9\sqrt{3}\)

Explanation: Like radical terms (same radicand) are combined by adding their coefficients. Both terms have \(\sqrt{3}\), so add coefficients: \(7+2=9\), giving \(9\sqrt{3}\). Option B is incorrect because it shows an incorrect coefficient (14); options C and D are wrong because they change the radicand (\(\sqrt{6}\) or \(\sqrt{30}\)), which cannot result from adding like radicals. Quick exam tip: combine radicals only when the radicands are identical; otherwise simplify each term first.

Correct answer: A. 12

Explanation: Compute directly: \(3\sqrt{2}\times 2\sqrt{2}=3\times2\times(\sqrt{2}\times\sqrt{2})=6\times2=12\). Option B (\(6\sqrt{2}\)) is incorrect because it ignores that \(\sqrt{2}\times\sqrt{2}=2\). Exam tip: simplify products of identical radicals using \(\sqrt{a}\times\sqrt{a}=a\).

Correct answer: A. 0

Explanation: The governing concept is simplifying surds by extracting perfect-square factors and then combining like radical terms. First, √12=√(4×3)=2√3, so 3√12=6√3. Next, √75=√(25×3)=5√3, so 2√75=10√3. Substituting these equivalent forms gives 4√3+6√3−10√3=(4+6−10)√3=0√3=0. Therefore option A is correct. Option B results from stopping after the first simplification or mishandling the subtraction. Option C is the positive value of the final subtracted term, not the complete expression. Option D comes from an incorrect combination of coefficients. Only A accounts for all three like-radical terms and their signs.

Correct answer: A. √48<√75<√108

Explanation: The governing property is that the square-root function is increasing on non-negative numbers. Therefore, whenever 0≤a<b, we have √a<√b. The radicands here satisfy 48<75<108, so applying this property immediately gives √48<√75<√108. The result can also be verified by simplifying: √48=√(16×3)=4√3, √75=√(25×3)=5√3, and √108=√(36×3)=6√3. Since √3 is positive, 4√3<5√3<6√3. Hence option A is correct. Option B reverses the complete order, while options C and D interchange two of the terms. No decimal approximation is necessary because the monotonicity rule gives an exact comparison.

Correct answer: A. (2)

Explanation: The governing concept is simplifying each surd by removing perfect-square factors and then multiplying. We have \sqrt{18}=\sqrt{9\cdot2}=3\sqrt{2} and \sqrt{8}=\sqrt{4 aimes2}=2\sqrt{2}. Therefore the bracket becomes 3\sqrt{2}-2\sqrt{2}=\sqrt{2}. Multiplying by the outside factor gives \sqrt{2}\cdot\sqrt{2}=2, because \sqrt{a}\sqrt{a}=a for a non-negative a. Thus option A is correct. Option D is the unsimplified product, while B and C arise from incorrect multiplication or failure to subtract like surd terms correctly. The result is rational even though the original expression contains irrational factors.

Correct answer: A. Both are rational real

Explanation: To find the zeroes, set p(x)=0: x²−16=0. This is a difference of two squares, so (x−4)(x+4)=0. Therefore x=4 or x=−4. Both roots are integers, and integers are rational numbers; they are also real because they lie on the real number line. Hence both zeroes are rational real numbers, as stated in option A. The presence of a square does not automatically make a root irrational: √16 simplifies to 4. Option B incorrectly leaves √16 unsimplified, option C would require a negative discriminant, and option D wrongly assigns different types to the two roots.

Correct answer: B. 0

Explanation: The governing idea is direct substitution together with the square-root identity (√a)²=a for every non-negative real number a. Since x=√3, replace x in the expression by √3: x²−3=(√3)²−3. Squaring the complete radical gives (√3)²=3, so the expression becomes 3−3=0. Therefore option B is correct. Option A would be obtained by treating the expression as a quotient of identical quantities, which it is not. Option C gives only x² and forgets the subtraction of 3, while option D adds instead of subtracting. The important step is to square the entire √3, not just manipulate the radicand incorrectly.

Correct answer: A. √2

Explanation: The key concept is simplifying a radical by extracting the perfect-square factor. Since 8 = 4 × 2, √8 = √(4 × 2) = √4 · √2 = 2√2. Substituting this into the expression gives x = 2√2 − √2 = √2. Therefore option A is correct. Option B is the simplified value of √8 before subtracting √2, so it represents an incomplete calculation. Option C incorrectly adds the coefficients, and option D treats the radical expression as an ordinary integer operation. Like terms containing the same radical can be combined just as algebraic like terms are combined.

Correct answer: A. 6√2

Explanation: Answer: A, 6√2. Simplify a square root by taking the largest perfect-square factor outside the radical. Write 72=36×2, where 36 is a perfect square. Then √72=√(36×2)=√36×√2=6√2. The factor 2 cannot be simplified further, so this is the simplest form. Option B is not equal: (8√2)^2=128, not 72. Option C, 3√8, is equivalent in value but not fully simplified, because √8=√(4×2)=2√2 and hence 3√8=6√2. Option D is too large; its square is 288. A useful memory cue is to remove perfect-square factors in pairs: 36 contributes 6 outside and leaves 2 inside. Never take the square root of 72 as 72's approximate value without simplifying the exact radical.