Irrational numbers and real numbers

Correct answer: B. (k=2)

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

Correct answer: A. Quadratic polynomial with real coefficients

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

Correct answer: A. (1,\sqrt{3})

Explanation: 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.

Correct answer: A. (0,\sqrt{5})

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

Correct answer: C. (x^4-10x^2+1=0)

Explanation: 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.

Correct answer: A. (0)

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

Correct answer: A. (0)

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

Correct answer: A. 0

Explanation: 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.

Correct answer: A. (0)

Explanation: ((\sqrt{3})^3=3\sqrt{3}), so (3\sqrt{3}-3\sqrt{3}=0). Simplifying powers is the key step in such questions.

Correct answer: C. Only (\sqrt{2}) is the sole irrational zero while coefficients stay rational

Explanation: In a quadratic with rational coefficients an irrational zero comes with its conjugate. In exams be suspicious of a lone irrational root.

Correct answer: B. (m) is a perfect square

Explanation: For a positive integer (m), (\sqrt{m}) is rational only when (m) is a perfect square. Identifying perfect squares is important in exams.

Correct answer: C. (n=98)

Explanation: (98) is not a perfect square, so (\sqrt{98}=7\sqrt{2}) is irrational. In exams extract perfect-square factors.

Correct answer: B. This is impossible

Explanation: Square roots of distinct primes are different irrationals and their sum cannot be rational. In exams do not assume independent radicals can combine to a rational number.

Correct answer: A. \(3\sqrt{2}\), irrational

Explanation: Since \(\sqrt{8}=\sqrt{4\cdot2}=2\sqrt{2}\), we have \(a=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\). Because \(\sqrt{2}\) is irrational and multiplying an irrational number by a nonzero rational (like 3) yields an irrational number, \(3\sqrt{2}\) is irrational. Option D (\(2\sqrt{2}\)) is the nearest distractor but is a wrong simplification; options C and B are incorrect because you cannot combine different radicals by placing them under one square root (\(\sqrt{2}+\sqrt{8}\neq\sqrt{10}\)) or obtain 10 by simple addition. Exam tip: always combine like radicals first and do not try to add under a single radical unless algebraically valid.

Correct answer: A. (\sqrt{3})

Explanation: (\sqrt{27}=3\sqrt{3}) and (\sqrt{12}=2\sqrt{3}), so the difference is (\sqrt{3}). Simplify first in exams.

Correct answer: A. \(4\)

Explanation: x and y are conjugates. Use the difference of squares: \(xy=(\sqrt{6})^2-(\sqrt{2})^2=6-2=4\). Option C simplifies to \(2\sqrt{12}=4\sqrt{3}\), which is not 4; options B and D result from incorrect operations (e.g., adding squares or mis-distributing). Exam tip: for expressions of the form \(a+b\) and \(a-b\), immediately apply \( (a+b)(a-b)=a^2-b^2\).

Correct answer: A. (\sqrt{5}+2)

Explanation: (\frac{1}{\sqrt{5}-2}\times\frac{\sqrt{5}+2}{\sqrt{5}+2}=\frac{\sqrt{5}+2}{5-4}=\sqrt{5}+2). Rationalise the denominator in exams.

Correct answer: A. (\sqrt{3}-1)

Explanation: (\frac{2}{\sqrt{3}+1}\times\frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{2(\sqrt{3}-1)}{2}=\sqrt{3}-1). The conjugate makes the denominator rational.

Correct answer: A. (1:2)

Explanation: Direct answer: option A, \(1:2\). Simplify \(y=\sqrt8=\sqrt{4\times2}=2\sqrt2\). Therefore \(x:y=\sqrt2:2\sqrt2\). The common positive factor \(\sqrt2\) cancels, leaving \(1:2\). Option A is correct. Option B reverses the ratio and would describe \(y:x\), not \(x:y\). Option C treats \(\sqrt8\) as four times \(\sqrt2\), which is incorrect. Option D is not a simplified ratio because it leaves unlike forms and does not represent the correct second term after simplification. A ratio can be divided by the same non-zero factor in both parts. Memory cue: simplify each radical before cancelling common factors.

Correct answer: C. (\sqrt{45})

Explanation: (\sqrt{45}=3\sqrt{5}), which is real and irrational. In exams do not treat the square root of a negative number as real.

Correct answer: A. \(-\sqrt{3}\)

Explanation: Compute first: \(x^2=(\sqrt{3})^2=3\). Substitute: \(2x^2-x-6=2\cdot3-\sqrt{3}-6=6-\sqrt{3}-6=-\sqrt{3}\). Hence A is correct. A common mistake is stopping at \(6-\sqrt{3}\) (option D) by neglecting to subtract the final 6. Exam tip: evaluate powers first, then combine constant terms carefully.

Correct answer: A. (0)

Explanation: Since (x-1=-\sqrt{5}), ((x-1)^2=5), so (x^2-2x-4=0). Isolate the irrational part and square in exams.

Correct answer: B. It is non-terminating recurring decimal

Explanation: The denominator contains (13), and after simplification the denominator is not made only of (2) and (5). In exams always check prime factors of the denominator.

Correct answer: A. (\sqrt{2}=-\frac{p}{q}), so (\sqrt{2}) would be rational which is impossible

Explanation: Since (-\frac{p}{q}) is rational, this would make (\sqrt{2}) rational which is false. In exams recognize the contradiction method.

Correct answer: A. (1), rational

Explanation: (ab=(7)^2-(4\sqrt{3})^2=49-48=1), so it is rational. In exams apply (a^2-b^2) for conjugate pairs.