Irrational numbers

Correct answer: A. √2, √7, π

Option A contains √2, √7, and π. Since 2 and 7 are not perfect squares, √2 and √7 are irrational. The constant π is also irrational, so every number in option A is irrational. In option B, √4 = 2, which is rational, so that set is not exclusively irrational. In option C, 0.25 = 1/4 is rational, despite √3 and √5 being irrational. In option D, 2/3 is rational, although π and √11 are irrational. Therefore option A is the only option containing only irrational numbers. The governing check is to identify roots of non-perfect squares, recognise π as irrational, and distinguish terminating decimals or ratios of integers, which are rational. A single rational member is enough to reject an entire option.

Correct answer: A. 8 + 2√15, irrational

Use the square-of-a-sum identity: x² = (√5 + √3)² = (√5)² + 2(√5)(√3) + (√3)² = 5 + 2√15 + 3 = 8 + 2√15. The number √15 is irrational because 15 is not a perfect square. Multiplying it by the nonzero rational number 2 remains irrational, and adding the rational number 8 cannot make it rational. If 8 + 2√15 were rational, subtracting 8 and dividing by 2 would make √15 rational, which is impossible. Thus option A is correct. Option B omits the cross term, option C confuses the product 5×3 with the expansion, and option D is not algebraically equal to x².

Correct answer: A. Real rational

For a quadratic ax² + bx + c, the discriminant D = b² − 4ac determines the nature of its zeroes. Here a = 1, b = −14, and c = 45, so D = (−14)² − 4(1)(45) = 196 − 180 = 16. Because D is positive, the zeroes are real and distinct. Also, √D = √16 = 4 is rational. By the quadratic formula, x = [14 ± 4]/2, giving x = 9 and x = 5. Thus both zeroes are real rational, so option A is correct. Option C would require D = 0, option D would require D < 0, and option B would apply if D were positive but not a perfect square. The explicit roots confirm the classification.

Correct answer: A. Real irrational

The nature of the zeroes is found from the discriminant D = b² − 4ac. For p(x) = x² − 14x + 38, a = 1, b = −14, and c = 38. Thus D = (−14)² − 4(1)(38) = 196 − 152 = 44. Because D is positive, the two zeroes are real and distinct. However, 44 is not a perfect square; √44 = 2√11 is irrational. The quadratic formula therefore gives x = (14 ± √44)/2 = 7 ± √11, so both zeroes are real irrational numbers. Hence option A is correct. Option B would require a perfect-square discriminant, option C would require D = 0, and option D would require D < 0.

Correct answer: A. x² − 10x + 4

A quadratic polynomial with rational coefficients that has 5 + √21 as a zero must also have its irrational conjugate 5 − √21 as the other zero. Their sum is (5 + √21) + (5 − √21) = 10, and their product is (5 + √21)(5 − √21) = 25 − 21 = 4. A monic quadratic with roots r and s is x² − (r + s)x + rs. Substitution gives x² − 10x + 4. Therefore option A is correct. Option B has the wrong sign for the sum, option C uses incorrect sum and product values, and option D has the correct x-coefficient but an incorrect constant term. The conjugate-root property is essential here.