Irrational numbers and real numbers

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.

The conjugate of (5+\sqrt{2}) is (5-\sqrt{2}). In exams changing the middle sign is the key idea of a conjugate.

(\alpha+\beta=14) and (\alpha\beta=43) so (\alpha^2+\beta^2=(14)^2-2(43)=110). In exams use (\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta).

(\frac{1}{2+\sqrt{7}}) equals (\frac{2-\sqrt{7}}{-3}=\frac{\sqrt{7}-2}{3}). So (x+\frac{1}{x}=2+\sqrt{7}+\frac{\sqrt{7}-2}{3}=\frac{4+4\sqrt{7}}{3}).

(\frac{1}{2+\sqrt{7}}=\frac{\sqrt{7}-2}{3}) so the total is (\frac{4+4\sqrt{7}}{3}). In exams rationalize the reciprocal first.

For the monic quadratic p(x) = x² − 2kx + 20, the product of the zeroes equals the constant term, 20. The stated zeroes have product (k + √5)(k − √5) = k² − 5 by the difference-of-squares identity. Equating these products gives k² − 5 = 20, so k² = 25 and k = ±5. Since the question asks for the positive value, k = 5, making option A correct. The sum provides an additional check: (k + √5) + (k − √5) = 2k, which agrees with the coefficient relation for x² − 2kx + 20. Options B, C, and √15 do not satisfy k² = 25 and therefore cannot produce the stated constant term.

(\sqrt{9+16}=5) while (\sqrt{9}+\sqrt{16}=7). In exams do not split addition inside a radical.

Since (9<15<16), (3<\sqrt{15}<4) and (\sqrt{15}) is irrational. In exams compare between squares.

The rational square root of a positive integer is an integer only when it is a perfect square. In exams identifying perfect squares is important.

Use the identity \((a-b)^2=a^2+b^2-2ab\). With \(a=\sqrt{11}\) and \(b=\sqrt{2}\), we get \(x^2=(\sqrt{11})^2+(\sqrt{2})^2-2\sqrt{11}\sqrt{2}=11+2-2\sqrt{22}=13-2\sqrt{22}\). Option B has the sign of the middle term wrong; option C has the constant term incorrect; option D incorrectly multiplies the terms. Exam tip: always expand using \((a-b)^2\) and simplify \(\sqrt{a}\sqrt{b}=\sqrt{ab}\) carefully.

The governing identity is the product of conjugate binomials: (a+b)(a−b) = a²−b². Here αβ = (5 + 2√6)(5 − 2√6). Taking a = 5 and b = 2√6 gives αβ = 5² − (2√6)² = 25 − 4×6 = 25 − 24 = 1. Therefore option A is correct. Option B is only 5² and ignores the contribution of the radical terms. Option C is the value of (2√6)², not the complete product, and option D has no basis in the identity. Although α and β individually contain irrational terms, they are conjugates, so the cross terms cancel and their product becomes the rational number 1.

((2+\sqrt{3})(2-\sqrt{3})=1), so the inverse is (2-\sqrt{3}). In exams check that the product is (1).

The discriminant is (196-160=36) so the zeroes are rational. The correct type should be real rational.

(\sqrt{72}=6\sqrt{2}), (\sqrt{50}=5\sqrt{2}) and (\sqrt{8}=2\sqrt{2}). Hence the value is (3\sqrt{2}).

If (\frac{s}{r}) were rational then (s=r\cdot\frac{s}{r}) would be rational which is false. In exams check the non-zero condition.

Multiplying by the conjugate gives numerator ((\sqrt{5}-\sqrt{3})^2=8-2\sqrt{15}) and denominator (2). So the value is (4-\sqrt{15}) and the correct simple option is A.

Multiplying by the conjugate gives (\frac{8-2\sqrt{15}}{2}=4-\sqrt{15}). In exams simplify the fraction at the end.

For polynomials with rational coefficients, the conjugate of an irrational root is also a root. The conjugate of \(3+\sqrt{8}\) is \(3-\sqrt{8}\). Their sum is \(6\) and product is \(9-8=1\). So the quadratic with these roots is \(x^2-(\text{sum})x+\text{product}=x^2-6x+1\). Option B has the wrong sign for the linear term, C has the wrong constant term, and D matches neither sum nor product. Exam tip: always use \(x^2-(r+s)x+rs\) to form the quadratic from roots r and s; simplify radicals first if helpful.

The principal square root is non-negative so (\sqrt{a^2}=|a|). In exams be careful when (a) is negative.

(x^2-y^2=(x-y)(x+y)=(2\sqrt{5})(2\sqrt{6})=4\sqrt{30}). In exams identities save long calculations.