Irrational numbers and real numbers

Correct answer: A. Sum 6√2, product 17

Explanation: For a quadratic ax²+bx+c with zeroes α and β, the governing relations are α+β=−b/a and αβ=c/a. In p(x)=x²−6√2x+17, the coefficients are a=1, b=−6√2 and c=17. Hence α+β=−(−6√2)/1=6√2, while αβ=17/1=17. Therefore option A is correct. Option B keeps the wrong sign for the sum. Option C swaps the coefficient and constant-term roles, and option D gives the product the wrong sign. The presence of an irrational coefficient does not change Vieta’s relations; they apply to quadratic polynomials regardless of whether the coefficients are rational or irrational.

Correct answer: A. 5 + 2√6

Explanation: 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.

Correct answer: A. 0

Explanation: 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.

Correct answer: A. 12 and 31

Explanation: Let α = 6 + √5 and β = 6 − √5. Their sum is α + β = (6 + √5) + (6 − √5) = 12 because the two radical terms cancel. Their product is αβ = (6 + √5)(6 − √5). Using (a+b)(a−b) = a²−b², this becomes 6² − (√5)² = 36 − 5 = 31. Therefore the required sum and product are 12 and 31, respectively, so option A is correct. Option B forgets to subtract 5 from 36. Option C uses only one constant term for the sum, and option D gives the difference of the conjugate numbers rather than their sum. The conjugate structure makes both calculations straightforward.

Correct answer: A. 1

Explanation: The governing identity is the product of conjugate binomials: (a+b)(a−b) = a²−b². Here αβ = (5 + 2√6)(5 − 2√6). Taking a = 5 and b = 2√6 gives αβ = 5² − (2√6)² = 25 − 4×6 = 25 − 24 = 1. Therefore option A is correct. Option B is only 5² and ignores the contribution of the radical terms. Option C is the value of (2√6)², not the complete product, and option D has no basis in the identity. Although α and β individually contain irrational terms, they are conjugates, so the cross terms cancel and their product becomes the rational number 1.

Correct answer: C. 3.86

Explanation: The governing idea is comparison of a rational number, an irrational number, and decimal candidates. First, 23/6 = 3.833333... . Also, since 3.87² = 14.9769 and 3.88² = 15.0544, √15 is approximately 3.873. Thus the required interval is approximately 3.833 < x < 3.873. Option C, 3.86, lies inside this interval. Option A, 3.82, is too small; options B and D, 3.90 and 4.00, are greater than √15. Therefore only option C satisfies both strict inequalities. The endpoints themselves would not be accepted even if listed, because the wording says greater than and less than.