Irrational numbers and real numbers

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.

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.

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

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

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

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

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.

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.

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

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 \( (a+b)(a-b)=a^2-b^2\).

(\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.

(\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.

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.

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

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.

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

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.

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

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