Irrational numbers and real numbers

The pair (5+\sqrt{5}) and (5-\sqrt{5}) has sum (10) and product (20) so it also fails. The pair (5+2) and (5-2) would be rational so none of the given options fits.

For roots with sum \(S=\alpha+\beta\) and product \(P=\alpha\beta\), the quadratic is \(x^2-Sx+P=0\). Here we get \(x^2-10x+21=0\). The discriminant is \(\Delta=10^2-4\cdot1\cdot21=16\). Thus the roots are \(x=\dfrac{10\pm\sqrt{16}}{2}=\dfrac{10\pm4}{2}\), giving 7 and 3. Option B is a tempting irrational pair but wrong since \((5+\sqrt{5})(5-\sqrt{5})=25-5=20\), not 21. Exam tip: form \(x^2-(\text{sum})x+(\text{product})=0\) and use \(\Delta\) to find roots quickly.

Real numbers include both rational and irrational numbers. In exams remember the inclusion of number systems.

A rational number has a terminating decimal expansion after it is reduced to lowest form only when the prime factors of its denominator are 2 and/or 5. The numerator is \(231=3\cdot7\cdot11\). In the denominator, \(2\cdot3\cdot5^2\cdot7\cdot11\) contains the same factors 3, 7, and 11, so they cancel with the numerator. The reduced fraction therefore has denominator \(2\cdot5^2=50\).

Since 50 has no prime factor other than 2 and 5, its decimal expansion terminates. In fact, the fraction becomes \(\frac{1}{50}=0.02\). Thus option A is correct. A recurring decimal would occur if another prime factor remained in the reduced denominator, but no such factor remains here. The supplied explanation correctly applies the terminating-decimal test.

(\sqrt{50}=5\sqrt{2}), (3\sqrt{8}=6\sqrt{2}) and (\sqrt{18}=3\sqrt{2}). Hence the value is (8\sqrt{2}).

(x^2=3+2\sqrt{2}) and (x^3=7+5\sqrt{2}), so (x^3-3x=4+2\sqrt{2}). The correct value is not in the options so calculate carefully.

(x^3=7+5\sqrt{2}) and (3x=3+3\sqrt{2}), so the difference is (4+2\sqrt{2}). In exams calculate powers step by step.

The sum of (9+\sqrt{7}) and (9-\sqrt{7}) is (18), and the product is (81-7=74). In exams check both sum and product of options.

Multiplying by the conjugate gives (\frac{30+2\sqrt{221}}{4}=\frac{15+\sqrt{221}}{2}). In exams divide by the common factor at the end.

After cancelling (385=5\cdot7\cdot11), only (2^3) remains in the denominator. In exams always check the denominator in lowest form.

The first product is (14-6=8), and (\sqrt{84}=2\sqrt{21}). In exams use both conjugate multiplication and radical simplification.

The other zero will be (6-\sqrt{19}), and the product is (36-19=17). In exams connect the constant term with the product of zeroes.

The governing concept is rationalisation by multiplying by the conjugate. Since p=6−√35, its reciprocal is 1/(6−√35). Multiply numerator and denominator by 6+√35: 1/p=(6+√35)/[(6−√35)(6+√35)]=(6+√35)/(36−35)=6+√35. Now subtract p: 1/p−p=(6+√35)−(6−√35)=2√35. Therefore option A is correct. Option B incorrectly cancels the radical terms, option C loses the factor 2 produced when subtracting a negative radical, and option D introduces an unsupported factor 6. The denominator is non-zero because √35 is less than 6, so the reciprocal is well defined.

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.

The governing idea is comparison of a rational number, an irrational number, and decimal candidates. First, 23/6 = 3.833333... . Also, since 3.87² = 14.9769 and 3.88² = 15.0544, √15 is approximately 3.873. Thus the required interval is approximately 3.833 < x < 3.873. Option C, 3.86, lies inside this interval. Option A, 3.82, is too small; options B and D, 3.90 and 4.00, are greater than √15. Therefore only option C satisfies both strict inequalities. The endpoints themselves would not be accepted even if listed, because the wording says greater than and less than.