Irrational numbers and real numbers

The sum is ((3+\sqrt{2})+(3-\sqrt{2})=6). In exams the sum of conjugate zeroes is always rational.

Verification:


i) (2+\sqrt{3})^2 = 2^2 + (\sqrt{3})^2 + 2\cdot2\cdot\sqrt{3} = 4 + 3 + 4\sqrt{3} = 7 + 4\sqrt{3}. Hence (2+\sqrt{3}) is the required number.

Why the others fail (brief):
- (3+\sqrt{2})^2 = 9 + 2 + 6\sqrt{2} = 11 + 6\sqrt{2}, not matching the given form.
- (\sqrt{7}+2)^2 = 7 + 4 + 4\sqrt{7} = 11 + 4\sqrt{7}, contains \sqrt{7} terms.
- (\sqrt{3}+1)^2 = 3 + 1 + 2\sqrt{3} = 4 + 2\sqrt{3}, middle term 2\sqrt{3} is too small.

Exam tip: compute the middle term 2ab to match the coefficient of the surd; that quickly eliminates wrong choices.

The square root of a perfect square is an integer, so for an irrational square root (m) is not a perfect square. In exams identifying perfect squares is important.

Rationalizing (\frac{1}{2-\sqrt{3}}) with (2+\sqrt{3}) gives (2+\sqrt{3}). In exams multiply by the conjugate of the denominator.

(4) is rational and (\sqrt{13}) is irrational, so the sum is irrational. In exams identify square roots of perfect squares first.

The sum is ((2+\sqrt{5})+(2-\sqrt{5})=4), which is rational. In exams remember conjugate pairs as counterexamples.

From (\sqrt{3}=\frac{p}{q}), we get (p^2=3q^2), so both (p) and (q) become divisible by (3). In exams use the coprime condition at the end.

The sum is (2) and the product is (1-6=-5), so the polynomial is (x^2-2x-5). In exams use (a^2-b^2) for the product.

Using the quadratic formula, (x=\frac{4\pm\sqrt{16+4}}{2}=2\pm\sqrt{5}). In exams simplify the discriminant.

(\sqrt{-9}) is not a real number, while the others are real. In exams do not take the square root of a negative number in the real number system.

(\frac{1}{3+\sqrt{10}}=\sqrt{10}-3), so the sum is (2\sqrt{10}). In exams rationalize the reciprocal first.

(\alpha+\beta=8) and (\alpha\beta=16-15=1), so the total is (9). In exams find the sum and product separately.

After simplification, (7) remains in the denominator, so the decimal is non-terminating recurring. In exams do not decide only from the original denominator.

(\sqrt{20}+\sqrt{45}=2\sqrt{5}+3\sqrt{5}=5\sqrt{5}), which is irrational. In exams do not treat addition like multiplication.

Use the identity \((a-b)^2=a^2+b^2-2ab)\). With \(a=\sqrt{7},\; b=\sqrt{3}\) we get \(x^2=(\sqrt{7}-\sqrt{3})^2=7+3-2\sqrt{21}=10-2\sqrt{21}\). Option B (\(10+2\sqrt{21}\)) is the typical sign-error from treating the cross term as positive. Exam tip: always expand using the identity and check the sign of the \(2ab\) term.

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.