Irrational numbers and real numbers

Core idea: compute \(p(\sqrt{7}+1)=(\sqrt{7}+1)^2-7\) by expanding. \((\sqrt{7}+1)^2=(\sqrt{7})^2+2\cdot\sqrt{7}\cdot1+1^2=7+2\sqrt{7}+1\). Subtracting 7 gives \(7+2\sqrt{7}+1-7=1+2\sqrt{7}\). Thus the correct value is \(1+2\sqrt{7}\). Closest distractor: option B (\(8+2\sqrt{7}\)) arises from forgetting to subtract the final \(7\). Option C (\(1\)) ignores the surd term \(2\sqrt{7}\). Option D (\(7+2\sqrt{7}\)) reflects a mis-evaluation of the constant terms. Exam tip: always expand \((a+b)^2\) carefully and simplify each term step by step to avoid dropping surd or constant terms.

Substitution gives (2+k\sqrt{2}-2=k\sqrt{2}=0), so (k=0). In exams write (p(\alpha)=0) for a zero.

(\frac{1}{1+\sqrt{2}}=\sqrt{2}-1), so the sum is (2\sqrt{2}). Rationalising the denominator is a quick exam method.

Because ((3+\sqrt{8})(3-\sqrt{8})=9-8=1), the reciprocal is (3-\sqrt{8}). If the product is (1), the reciprocal is immediate.

(\alpha\beta=(\sqrt{5}+2)(\sqrt{5}-2)=5-4=1). Conjugate multiplication often gives a rational answer.

The sum is (10) and the product is (25-6=19). In exams quickly find sum and product for conjugate zeroes.

The discriminant is (100-76=24), so the zeroes are (\frac{10\pm\sqrt{24}}{2}=5\pm\sqrt{6}). Simplify (\sqrt{24}=2\sqrt{6}) in exams.

The discriminant is (36-4=32), positive but not a perfect square. So there are two distinct irrational real roots.

The discriminant is (16-16=0), so the roots are equal and (x=2) is rational. In exams (D=0) means equal roots.

The discriminant is (4-20=-16), which is negative, so there are no real roots. In exams do not treat a negative discriminant as real zeroes.

For (k=2), the discriminant is (16-8=8), positive but not a perfect square. Therefore the roots are real and irrational.

(\sqrt{2}) is real but irrational, so the coefficients are real but not all rational. Since the degree is (2), it is quadratic.

The sum is (1+\sqrt{3}) and the product is (\sqrt{3}), so the zeroes are (1) and (\sqrt{3}). Compare with (x^2-Sx+P) in exams.

(p(x)=x(x-\sqrt{5})), so the zeroes are (0) and (\sqrt{5}). Taking the common factor is a fast method in exams.

Answer: A, 5 + 2√6. Use the identity (a + b)² = a² + 2ab + b². Put a = √2 and b = √3. Then x² = (√2 + √3)² = (√2)² + 2(√2)(√3) + (√3)². The first square is 2 and the last square is 3. For the middle product, √2·√3 = √(2·3) = √6, so the cross-term is 2√6. Hence x² = 2 + 2√6 + 3 = 5 + 2√6. A is therefore correct. B omits the factor 2 in the cross-term, which is the most common error. C wrongly changes √2·√3 into √5 and does not write the complete expansion. D uses incorrect radicands in both the constant and radical parts. Memory cue: when a sum is squared, never forget 2ab; the cross-term is essential.

Here (x^2=5+2\sqrt{6}) and then ((x^2-5)^2=24), so (x^4-10x^2+1=0). In exams a sum of two radicals may lead to a fourth-degree relation.

Since (x-2=\sqrt{5}), squaring gives ((x-2)^2=5), so (x^2-4x-1=0). In exams square to remove the radical.

Since (x-3=-\sqrt{2}), ((x-3)^2=2), hence (x^2-6x+7=0). Handle (a-\sqrt{b}) the same way in exams.

Since \(x^2=11\), we have \(x^4=(x^2)^2=11^2=121\). Therefore \(x^4-121=121-121=0\). Option C (121) is the value of \(x^4\) alone; the expression subtracts 121, so it cancels to 0. Exam tip: reduce higher powers stepwise (use \(x^2\) first) before substituting and simplifying.

The governing concept is evaluation of a polynomial by substitution. Replace x with √2 in p(x) = x³ − 2x: p(√2) = (√2)³ − 2√2. Since (√2)³ = (√2)²√2 = 2√2, the expression becomes p(√2) = 2√2 − 2√2 = 0. Therefore option A is correct. The same result can be checked by factoring p(x) = x(x² − 2). At x = √2, the factor x² − 2 becomes 2 − 2 = 0, so the entire product is zero. Option B represents only the second term without its subtraction, option C is only the first term, and option D does not result from the substitution.