Irrational numbers and real numbers

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.

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.

The number halfway between two different real numbers is their average. If the numbers are
\(\sqrt{2}\) and \(\sqrt{3}\), their average is \(\frac{\sqrt{2}+\sqrt{3}}{2}\). An average of two numbers lies strictly between them when the numbers are unequal. Therefore option A gives a number between \(\sqrt{2}\) and \(\sqrt{3}\). The other options simplify to 1, 1, and \(\frac12\), respectively, so they are not between these two values.

To verify this numerically, \(\sqrt2\approx1.414\) and \(\sqrt3\approx1.732\). Their average is approximately \(1.573\), which is greater than 1.414 and less than 1.732. Thus the stated answer A is correct. The key idea is that the midpoint of an interval always lies inside that interval, not at either endpoint.

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.

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.

Compute each term separately: \(\sqrt{0.49}=0.7\) and \(\frac{1}{\sqrt{25}}=\frac{1}{5}=0.2\). Their sum is \(0.7+0.2=0.9\). Option B (0.7) represents only \(\sqrt{0.49}\) and omits the second term, so it is incorrect. Exam tip: simplify square roots first (including decimals), then perform the arithmetic to avoid sign or omission errors.

(\sqrt{4n}=2\sqrt{n}). Multiplying an irrational number by a non zero rational keeps it irrational.

(\sqrt{8}=2\sqrt{2}), so the sum is (3\sqrt{2}). A non zero rational multiple of (\sqrt{2}) remains irrational.

(\sqrt{98}=7\sqrt{2}) and (\sqrt{18}=3\sqrt{2}). The numerator is (4\sqrt{2}), and division gives (4).

Multiplying by the conjugate makes the denominator (1). The numerator is \((\sqrt{3}+\sqrt{2})^2=5+2\sqrt{6}\).

Multiplying by the conjugate gives denominator (5-3=2) and numerator (8+2\sqrt{15}). The simplified form is (4+\sqrt{15}).

Write each radical in simplest form: \sqrt{12}=2\sqrt{3}, \sqrt{27}=3\sqrt{3}, \sqrt{75}=5\sqrt{3}, \sqrt{48}=4\sqrt{3}. Adding and subtracting gives 2\sqrt{3}+3\sqrt{3}+5\sqrt{3}-4\sqrt{3}=6\sqrt{3}, so A is correct. Options B and C reflect common arithmetic mistakes in combining coefficients; option D (\sqrt{66}) is not possible here because you must combine like radicals (same radicand) after simplifying. Exam tip: factor out perfect square factors first, then combine like radical terms by adding their coefficients.

The decimal expansion is infinite and the gaps of zeros between successive 1s increase: patterns of 1 appear at positions 1, 3, 6, 10, ... (triangular numbers). A rational number must have a decimal expansion that eventually becomes periodic (a fixed repeating block). Since no such periodic block exists here, the number is irrational. The closest distractor, rational number, is incorrect because there is no eventual repetition; terminating decimal is wrong because the expansion does not end; integer is impossible because the value lies between 0 and 1. Exam tip: to test for irrationality, look for eventual periodicity — if none exists and the expansion is infinite, the number is irrational.

The square root of a positive integer is rational only when it is a perfect square. This is the key rule for roots.

(\sqrt{289}=17) is rational and (\sqrt{290}) is irrational. A rational plus an irrational is irrational.

\(\sqrt[3]{216}=6\) and \(\sqrt[3]{125}=5\) because 216 and 125 are perfect cubes (6^3=216, 5^3=125). Therefore the sum is \(6+5=11\). Option D arises if one mistakenly takes the cube root after adding (\(\sqrt[3]{216+125}=\sqrt[3]{341}\) ), which is not the same; its value is about 6.99. Exam tip: identify perfect cubes first, take each cube root, then perform the required arithmetic—don’t swap root and sum operations.

Since \(z=\sqrt[3]{9}\), we have \(z^3=(\sqrt[3]{9})^3=9\). The number 9 is an integer and therefore rational (9 = 9/1). Thus \(z^3\) is rational. Option B is wrong because 9 is not irrational; C is wrong because 9 is a real number (not non‑real); D is wrong because 9 is algebraic (it satisfies \(x-9=0\)), so it is not transcendental. Exam tip: simplify powers and roots first — often the result is an obvious integer or rational number after simplification.

(180) is not a perfect square so (\sqrt{180}) is irrational. Subtracting an irrational from a rational gives an irrational result.

(\sqrt{8}=2\sqrt{2}), so (\sqrt{2}+\sqrt{8}=3\sqrt{2}). The unlike root (\sqrt{5}) remains separate.

Write \(\sqrt{20}=2\sqrt{5}\). Then \((\sqrt{20}-\sqrt{5})=2\sqrt{5}-\sqrt{5}=\sqrt{5}\), and its square is \((\sqrt{5})^2=5\). Using the binomial formula \((a-b)^2=a^2+b^2-2ab\) gives \(20+5-2\sqrt{20}\sqrt{5}=25-20=5\). Option B (15) arises from wrongly doing \(20-5\) and ignoring the middle term; option C (\(\sqrt{15}\)) mistakes squaring for taking a square root of the difference. Exam tip: simplify surds first, then apply the formula for \((a-b)^2\).