Irrational numbers and real numbers

The companion zero is (2-\sqrt{3}), so the factor is (x-(2-\sqrt{3})). In exams remember the relation between a zero and factor as (x-\alpha).

(\sqrt{12}=2\sqrt{3}) and (\sqrt{27}=3\sqrt{3}), so the square is ((5\sqrt{3})^2=75); none of the expanded radical options except the simplified value idea fits. In exams simplify before expanding.

(\sqrt{12}+\sqrt{27}=2\sqrt{3}+3\sqrt{3}=5\sqrt{3}), so the square is (75). In exams add like radicals first.

The sum is (12) and the product is (36-11=25), so the polynomial is (x^2-12x+25). In exams write the standard form correctly.

The conjugate of the denominator is (\sqrt{7}-\sqrt{6}), and the denominator becomes (7-6=1). In exams the answer simplifies when the difference is (1).

(\sqrt{8}+(-\sqrt{8})=0), which is rational. In exams one counterexample is enough to disprove a universal statement.

If (q\ne0), then (q\sqrt{5}) is irrational and the sum cannot be rational. In exams check the possibility of a zero coefficient.

Multiplying by the conjugate of the denominator gives denominator (1) and numerator ((\sqrt{3}+\sqrt{2})^2=5+2\sqrt{6}). In exams apply the conjugate in one step.

From the product (k^2-7=9), we get (k^2=16), and (k=4) fits the given form. In exams use the product to find the unknown.

(\sqrt{a+b}=\sqrt{a}+\sqrt{b}) is generally false, for example (\sqrt{9+16}\ne3+4). In exams do not split addition inside a radical.

Since (2<\frac{5}{2}<3), (\sqrt{2}<\sqrt{\frac{5}{2}}<\sqrt{3}), and it is irrational. In exams compare by squaring.

The companion zero is (5-2\sqrt{6}), with sum (10) and product (25-24=1). In exams form the polynomial using the conjugate.

The product ((3+\sqrt{2})(3-\sqrt{2})=9-2=7), so (k=7). In exams connect the constant term with the product of zeroes.

Multiplying by the conjugate (\sqrt{11}+3) makes the denominator (11-9=2), and (2) cancels. In exams choose the conjugate of the denominator correctly.

Since ((\sqrt{13}+\sqrt{12})(\sqrt{13}-\sqrt{12})=1), the reciprocal is (\sqrt{13}-\sqrt{12}). In exams quickly identify conjugates where (a-b=1).

(154=2\cdot7\cdot11), so after cancellation only (5^2) remains in the denominator. In exams decide from the denominator in lowest form.

(\sqrt{12}=2\sqrt{3}), so (\sqrt{3}+\sqrt{12}=3\sqrt{3}). In exams make radicals like terms before adding.

With rational coefficients (a+\sqrt{b}) is accompanied by (a-\sqrt{b}). In exams identify conjugate zeroes quickly.

The sum of zeros is \((2+\sqrt{10})+(2-\sqrt{10})=4\) and the product is \((2+\sqrt{10})(2-\sqrt{10})=4-10=-6\). For a monic quadratic the polynomial is \(x^2-(\text{sum})x+\text{product}\), so \(x^2-4x-6\) is correct. Closest distractors fail because: \(x^2-4x+6\) has the wrong constant term (+6 instead of -6), \(x^2+4x-6\) has the wrong sign for the linear term, and \(x^2-2x-10\) has neither sum nor product matching. Exam tip: compute sum and product first and form \(x^2-({\text{sum}})x+{\text{product}}\).

Core concept: use \((a+b)^2=a^2+b^2+2ab\). With \(a=\sqrt{7},\; b=\sqrt{5}\) we get \(x^2=(\sqrt{7})^2+(\sqrt{5})^2+2\sqrt{7}\sqrt{5}=7+5+2\sqrt{35}=12+2\sqrt{35}\). Option B is a close distractor that uses the radical but with the wrong coefficient (misses the factor 2); option C omits the cross term entirely. Exam tip: always include the \(2ab\) term and simplify \(\sqrt{m}\sqrt{n}=\sqrt{mn}\).

((\sqrt{28})(\sqrt{7})=\sqrt{196}=14) which is rational. In exams keep multiplication and addition rules separate.

(-\frac{p}{q}) is rational so it would make (\sqrt{3}) rational which is false. In exams recognize the contradiction method.

Multiply the conjugates: \(ab=(\sqrt{13}+\sqrt{6})(\sqrt{13}-\sqrt{6})=(\sqrt{13})^2-(\sqrt{6})^2=13-6=7\). Choice B (19) is the sum \(13+6\), not the product of conjugates; choice C (\(\sqrt{78}\)) arises from incorrectly taking \(\sqrt{13\cdot6}\); choice D is a sign error. Exam tip: recognise conjugates and apply \(a^2-b^2\) to remove radicals quickly.

The conjugate of the denominator is (\sqrt{13}+2) and the denominator becomes (13-4=9). Hence the value is (\frac{3(\sqrt{13}+2)}{9}=\frac{\sqrt{13}+2}{3}).

Take reciprocal by rationalizing denominator using the conjugate. \(\dfrac{1}{4-\sqrt{15}}=\dfrac{4+\sqrt{15}}{(4-\sqrt{15})(4+\sqrt{15})}=\dfrac{4+\sqrt{15}}{4^2-(\sqrt{15})^2}=\dfrac{4+\sqrt{15}}{16-15}=4+\sqrt{15}\). Thus the reciprocal is \(4+\sqrt{15}\). Option C is a common arithmetic error where someone adds squares to get \(31\) (i.e. uses \(16+15\)) instead of using the difference \(a^2-b^2\). Option B is just the original number, not its reciprocal. Exam tip: multiply numerator and denominator by the conjugate and use \(a^2-b^2\) to simplify quickly.