Irrational numbers and real numbers

The conjugate of (\sqrt{11}=0+\sqrt{11}) is (-\sqrt{11}). In exams also identify the case (a=0).

The discriminant is (100-92=8), and (\sqrt{8}) is irrational, so the zeroes are real irrational. In exams check the square root of the discriminant.

(\sqrt{50}=5\sqrt{2}) and (\sqrt{32}=4\sqrt{2}), so the value is (2\sqrt{2}). In exams handle signs carefully.

Since (9<13<16), (3<\sqrt{13}<4), and (\sqrt{13}) is irrational. In exams compare between squares.

The sum of (4+\sqrt{5}) and (4-\sqrt{5}) is (8), and the product is (16-5=11). In exams check the sum and product of options.

The principal square root is always non-negative, so (\sqrt{a^2}=|a|). In exams do not forget the possibility of negative (a).

(x^2-y^2=(x-y)(x+y)=(2\sqrt{2})(2\sqrt{5})=4\sqrt{10}). In exams use identities to avoid long calculation.

(\sqrt{17}) is irrational, so its decimal expansion is non-terminating non-recurring. In exams distinguish irrational decimals from recurring decimals.

The like (x) terms cancel and the value left is (2\sqrt{2}). In exams do not be confused by the type of number during algebraic simplification.

(\sqrt{2}\cdot3\sqrt{2}=6), which is rational. In exams remember counterexamples for products of irrational 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.