Irrational numbers and real numbers

The sum is (6) and the product is (7), so the polynomial is (x^2-6x+7). In exams use (x^2-(\text{sum})x+\text{product}).

Expand the square: x^2 = (\sqrt{6}+\sqrt{2})^2 = 6 + 2 + 2\sqrt{12} = 8 + 2\times 2\sqrt{3} = 8 + 4\sqrt{3}. Option B (8+2\sqrt{3}) is a common error caused by not fully simplifying \sqrt{12} (since \sqrt{12}=2\sqrt{3}, the cross-term becomes 4\sqrt{3}, not 2\sqrt{3}). Exam tip: apply (a+b)^2 = a^2+2ab+b^2 and simplify radicals fully before choosing the answer.

The denominator contains (7), so the decimal will not terminate and being rational it will recur. In exams decide after checking the denominator in lowest form.

In (\frac{63}{2^5\cdot5^2\cdot7}), after cancelling (63) and (7), only (2) and (5) remain in the denominator. In exams reduce the fraction first.

These are conjugate terms. \(ab=(\sqrt{11}+\sqrt{5})(\sqrt{11}-\sqrt{5})=(\sqrt{11})^2-(\sqrt{5})^2=11-5=6\). A common wrong choice is \(\sqrt{55}\), which equals \(\sqrt{11}\cdot\sqrt{5}\) but is not the product of the conjugates. Option 16 arises from adding 11 and 5 (not correct for multiplication), and \(2\sqrt{55}\) is an incorrect doubled product. Exam tip: recognize conjugates and apply the difference-of-squares identity \(a^2-b^2\) to remove radicals quickly.

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.