Irrational numbers and real numbers

\(\sqrt[3]{125}=5\), which is an integer and can be written as \(5=5/1\); therefore it is a rational number. The cube root of a perfect cube is an integer (hence rational). The closest distractor B (irrational) is incorrect because irrational numbers have non‑terminating, non‑repeating decimal expansions, whereas 5 is a terminating decimal. C is wrong because the number is real, not non‑real. D is incorrect because a "non‑repeating decimal" typically describes irrationals, but 5 has a terminating decimal. Exam tip: first check whether the radicand is a perfect power—if so, the root will be an integer (rational).

If \(\sqrt[3]{12}\) were rational it could be written as a reduced fraction \(\dfrac{p}{q}\) with integers p,q; cubing gives \(p^3=12q^3\). That forces the prime exponents in \(p^3\) to match those of 12q^3, so the exponents of primes coming from 12 must be multiples of 3. But 12=2^2·3^1 has exponents 2 and 1, which are not multiples of 3, a contradiction. Hence \(\sqrt[3]{12}\) is irrational. Distractors fail: 3 and 4 are wrong because 3^3=27 and 4^3=64 (not 12); a terminating decimal would be rational. Exam tip: first check whether the radicand is a perfect cube; use prime-power exponents to test rationality of integer roots.

(45) is not a perfect square so (\sqrt{45}) is irrational. Adding rational (2) keeps it irrational.

(\sqrt{100}=10) is rational so (9-\sqrt{100}) will be rational. The root of a perfect square is rational.

Multiply as conjugates using the identity \\(a+b)(a-b)=a^2-b^2\\. With a=1 and b=\sqrt{7} we get \\(1+\sqrt{7})(1-\sqrt{7})=1^2-(\sqrt{7})^2=1-7=-6\\. Option B (6) is a sign error; C (0) and D (-8) are results of arithmetic mistakes. Exam tip: for conjugate pairs use the difference of squares formula directly to avoid algebraic expansion errors.

(\sqrt{48}=4\sqrt{3}), (\sqrt{108}=6\sqrt{3}), and (\sqrt{12}=2\sqrt{3}). The result is (8\sqrt{3}).

The governing ideas are conjugate surds and the difference-of-squares identity. Add the expressions term by term: (2+√3)+(2−√3) = 2+2+√3−√3 = 4, because the radical terms cancel. For the product, use (a+b)(a−b)=a²−b². Thus (2+√3)(2−√3) = 2²−(√3)² = 4−3 = 1. Therefore the correct pair is sum 4 and product 1, given in option A. Option B incorrectly cancels the constant terms as well. Option C gives the wrong sum and product, and option D gets the sum right but incorrectly evaluates the product as 7. The conjugate structure makes the calculation especially direct.

Let \(a=\sqrt{5}\) and \(b=\sqrt{2}\). Using \((a-b)^2=a^2+b^2-2ab\) we get \((\sqrt{5}-\sqrt{2})^2=5+2-2\sqrt{10}=7-2\sqrt{10}\). So option A is correct. Option B misses the factor 2 in the cross term; option C has the wrong sign for the cross term; option D omits the cross term entirely. Exam tip: always apply the \(2ab\) term with the correct sign when expanding squares of binomials.

A rational number has a decimal expansion that either ends after a finite number of digits or continues in a repeating pattern. For example, \(1/4=0.25\) terminates, while \(1/3=0.333\ldots\) repeats. This gives a useful way to identify rational numbers from their decimal forms. The repeating block may contain one digit or several digits.

The statement says that the decimal is terminating or repeating, so it matches the defining decimal property of rational numbers. An irrational number has a decimal expansion that is non-terminating and non-repeating. Integers are only a small group of rational numbers, so not every rational number is an integer. Therefore option A, rational number, is correct.

\(\sqrt{11}\) is irrational while \(\frac{1}{3}\) is rational. Assume for contradiction that \(\frac{1}{3}+\sqrt{11}\) is rational; then \(\sqrt{11}=\big(\frac{1}{3}+\sqrt{11}\big)-\frac{1}{3}\) would be a difference of two rationals and hence rational, contradicting that \(\sqrt{11}\) is irrational. Thus the sum is irrational. The closest distractor, “rational,” fails because a rational plus an irrational cannot be rational. Exam tip: when one term is irrational, directly use contradiction (subtract the rational term) to show the sum is irrational.

\(\sqrt{112}=\sqrt{16\times7}=4\sqrt{7}\). Hence \(4\sqrt{7}-\sqrt{112}=4\sqrt{7}-4\sqrt{7}=0\). Option A is correct. A tempting distractor is \(\sqrt{7}\), but it is not equal because the subtraction yields zero after simplifying \(\sqrt{112}\). Exam tip: always factor out perfect square factors from square roots to simplify surds quickly.

Factor 245: \(245=49\times5\). Then \(\sqrt{245}=\sqrt{49\times5}=\sqrt{49}\times\sqrt{5}=7\sqrt{5}\), so A is correct. Option B (\(5\sqrt{7}\)) is incorrect — \((5\sqrt{7})^2=175\), not 245. Option C (\(49\sqrt{5}\)) arises from mistakenly taking 49 instead of its square root (\(\sqrt{49}=7\)), giving an answer that is far too large. Option D is just the unsimplified radical. Exam tip: always pull out the largest perfect square factor and take its square root outside the radical.

The direct answer is option A: 9. This is a difference of squares: \\( (a+b)(a-b)=a^2-b^2\\). Take \\(a=4\\) and \\(b=\sqrt7\\). Then \\( (4+\sqrt7)(4-\sqrt7)=4^2-(\sqrt7)^2=16-7=9\\), because \\( (\sqrt7)^2=7\\). Option A is correct. Option B, 23, comes from adding 16 and 7, but conjugate multiplication requires subtraction. Option C, \\(16+7=23\\), is exactly that same incorrect addition. Option D, \\(8\sqrt7\\), would arise from keeping only the two cross-like terms and ignoring the square terms; in fact those middle terms cancel. The conjugate pair is useful because the radical disappears. Memory cue: plus-minus pairs use “first square minus second square.”

(\sqrt{72}=6\sqrt{2}), (\sqrt{128}=8\sqrt{2}), and (\sqrt{50}=5\sqrt{2}). The result is (9\sqrt{2}).

Since (64<70<81), (\sqrt{70}) lies between (8) and (9). (70) is not a perfect square so it is irrational.

On subtracting, the (2) terms cancel and (\sqrt{10}-(-\sqrt{10})=2\sqrt{10}). Watch the signs carefully.

Use the identity \((a-b)(a+b)=a^2-b^2\). With \(a=\sqrt{13}\) and \(b=\sqrt{3}\) we get \(13-3=10\), so the value is 10. Option C (\(\sqrt{39}\)) confuses the product of the individual square roots with the whole conjugate product; \(\sqrt{13}\cdot\sqrt{3}=\sqrt{39}\) is not equal to \((\sqrt{13}-\sqrt{3})(\sqrt{13}+\sqrt{3})\). Exam tip: spot the conjugate pair and apply the difference-of-squares formula to simplify quickly.

\( (\sqrt{6}+\sqrt{2})^2 = (\sqrt{6})^2 + 2\cdot\sqrt{6}\cdot\sqrt{2} + (\sqrt{2})^2 = 6 + 2\sqrt{12} + 2.\) Since \(\sqrt{12}=2\sqrt{3}\), the middle term becomes \(2\sqrt{12}=4\sqrt{3}\). Thus the expansion is \(8+4\sqrt{3}\). Option B (\(8+2\sqrt{3}\)) reflects the common mistake of treating \(\sqrt{12}\) as \(\sqrt{3}\), so it is incorrect. Exam tip: apply \((a+b)^2=a^2+2ab+b^2\) and simplify radicals by extracting perfect-square factors.

\(\sqrt[3]{216}=6\) because \(216=6^3\). Since 6 is an integer, it is a rational number (can be written as p/q). Options B and D are descriptions of irrational numbers (whose decimal expansion is non-terminating and non-repeating), so they are wrong. Option C is incorrect because non-real numbers have an imaginary component; here the cube root is a real rational number. Exam tip: when finding a cube root, first check whether the number is a perfect cube — if so, the cube root is an integer.

(\sqrt{121}=11) is rational so (5+\sqrt{121}) will be rational. The root of a perfect square is rational.

\(\sqrt{288}=\sqrt{144\times2}=\sqrt{144}\times\sqrt{2}=12\sqrt{2}\), so the simplified form is \(12\sqrt{2}\). The closest distractor \(2\sqrt{144}\) equals \(2\times12=24\), which is not equal to \(12\sqrt{2}\). Exam tip: always factor the radicand into the largest perfect square times a remaining factor to simplify surds quickly (here the largest perfect square is 144).

The governing principle is to extract perfect-square factors and then combine like surds. Since 63 = 9 × 7, we have √63 = √(9 × 7) = 3√7. Substituting this into the expression gives √7 + √63 = √7 + 3√7 = 4√7. The two terms are like surds because both contain √7, so their coefficients, 1 and 3, can be added. Therefore option A is correct. Option B uses an incorrect coefficient of 10. Option C treats addition as though it were multiplication and changes the expression improperly. Option D changes the radicand and does not preserve the original value. Exact radical simplification is preferable to using a decimal approximation.

Simplify each radical: \(\sqrt{98}=\sqrt{49\cdot2}=7\sqrt{2}\) and \(\sqrt{8}=\sqrt{4\cdot2}=2\sqrt{2}\). Their difference is \(7\sqrt{2}-2\sqrt{2}=5\sqrt{2}\), so option A is correct. Distractors like C (\(3\sqrt{2}\)) and B (\(9\sqrt{2}\)) are incorrect because the coefficients are not the result of subtracting 7 and 2. Exam tip: extract square factors first to write surds as a common radical factor, then subtract the coefficients.

The direct answer is A: \(\sqrt{18}\times\sqrt{50}=30\). Apply \(\sqrt{x}\sqrt{y}=\sqrt{xy}\) for nonnegative radicands. Then \(\sqrt{18}\times\sqrt{50}=\sqrt{18\cdot50}=\sqrt{900}=30\), because \(30^2=900\). Option A is therefore correct. Option B, \(\sqrt{68}\), comes from adding 18 and 50 instead of multiplying them, so it does not represent the product. Option C, \(15\sqrt2\), is not the correct simplification; although \(\sqrt{18}=3\sqrt2\) and \(\sqrt{50}=5\sqrt2\), their product is \(15(\sqrt2)^2=15\cdot2=30\). Option D incorrectly multiplies by 50 while leaving \(\sqrt{18}\) unchanged. Notice that both factors are irrational, but their product is rational because the remaining square-root factors pair to make 2. The safest method is to multiply under one radical and then identify the perfect square. Do not add radicands when the radicals are being multiplied.

This is a product of conjugates — use the identity \((a+b)(a-b)=a^2-b^2\). With \(a=\sqrt{5}\) and \(b=\sqrt{2}\) we get \(a^2-b^2=5-2=3\). Option C (\(\sqrt{10}\)) confuses the product of the radicals \(\sqrt{5}\cdot\sqrt{2}=\sqrt{10}\) with the whole conjugate product; the middle terms cancel in the conjugate product, so the result is the rational number 3. Option B (\(-3\)) is a sign error. Exam tip: spot conjugates quickly and apply \((a+b)(a-b)=a^2-b^2\) to avoid expanding term-by-term.