Irrational numbers and real numbers

((2+\sqrt{3})(2-\sqrt{3})=4-3=1) which is rational. In exams remember conjugate multiplication as a counterexample.

From (\sqrt{5}=\frac{p}{q}) we get (p^2=5q^2) so both (p) and (q) are divisible by (5). This contradicts the coprime condition.

Using the quadratic formula (x=\frac{8\pm\sqrt{64-52}}{2}=4\pm\sqrt{3}). In exams simplify the discriminant.

(\sqrt{27}=3\sqrt{3}), (\sqrt{75}=5\sqrt{3}) and (\sqrt{12}=2\sqrt{3}). Hence the value is (6\sqrt{3}).

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.

The direct answer is option A: take \(a=9\) and \(b=16\). The claimed rule would say \(\sqrt{9+16}=\sqrt9+\sqrt{16}\). The left side is \(\sqrt{25}=5\), while the right side is \(3+4=7\). Since 5 and 7 are different, this is a genuine counterexample. Option B gives \(\sqrt{0+16}=4\) and \(\sqrt0+\sqrt{16}=0+4=4\), so the statement happens to work there. Option C gives \(\sqrt{9+0}=3\) and \(3+0=3\), so it also does not disprove the rule. Option D gives \(\sqrt0=0\) on both sides. Thus only A shows failure. The correct general rule is that square roots do not usually distribute over addition. Do not split a sum 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.

Direct answer: option A, \(2-\sqrt3\). The multiplicative inverse of a non-zero number is the number that gives 1 when multiplied by it. Multiply by the conjugate: \((2+\sqrt3)(2-\sqrt3)=2^2-(\sqrt3)^2=4-3=1\). Hence the inverse is \(2-\sqrt3\). Option A is correct. Option B is the original number, not its inverse; its product with itself is generally not 1. Option C is \(-(2-\sqrt3)\), so its product is -1. Option D divides the correct inverse by 7, making the product \(1/7\), not 1. The conjugate is useful because it changes a difference of surd squares into an ordinary number. Memory cue: test an inverse by multiplying; the result must be exactly 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.

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.