Irrational numbers and real numbers

The square root of a positive integer is rational only when it is a perfect square. This is the key rule for roots.

(\sqrt{289}=17) is rational and (\sqrt{290}) is irrational. A rational plus an irrational is irrational.

\(\sqrt[3]{216}=6\) and \(\sqrt[3]{125}=5\) because 216 and 125 are perfect cubes (6^3=216, 5^3=125). Therefore the sum is \(6+5=11\). Option D arises if one mistakenly takes the cube root after adding (\(\sqrt[3]{216+125}=\sqrt[3]{341}\) ), which is not the same; its value is about 6.99. Exam tip: identify perfect cubes first, take each cube root, then perform the required arithmetic—don’t swap root and sum operations.

Since \(z=\sqrt[3]{9}\), we have \(z^3=(\sqrt[3]{9})^3=9\). The number 9 is an integer and therefore rational (9 = 9/1). Thus \(z^3\) is rational. Option B is wrong because 9 is not irrational; C is wrong because 9 is a real number (not non‑real); D is wrong because 9 is algebraic (it satisfies \(x-9=0\)), so it is not transcendental. Exam tip: simplify powers and roots first — often the result is an obvious integer or rational number after simplification.

(180) is not a perfect square so (\sqrt{180}) is irrational. Subtracting an irrational from a rational gives an irrational result.

(\sqrt{8}=2\sqrt{2}), so (\sqrt{2}+\sqrt{8}=3\sqrt{2}). The unlike root (\sqrt{5}) remains separate.

Write \(\sqrt{20}=2\sqrt{5}\). Then \((\sqrt{20}-\sqrt{5})=2\sqrt{5}-\sqrt{5}=\sqrt{5}\), and its square is \((\sqrt{5})^2=5\). Using the binomial formula \((a-b)^2=a^2+b^2-2ab\) gives \(20+5-2\sqrt{20}\sqrt{5}=25-20=5\). Option B (15) arises from wrongly doing \(20-5\) and ignoring the middle term; option C (\(\sqrt{15}\)) mistakes squaring for taking a square root of the difference. Exam tip: simplify surds first, then apply the formula for \((a-b)^2\).

Multiply the conjugates using the difference of squares: \((\sqrt{7}+\sqrt{2})(\sqrt{7}-\sqrt{2})=(\sqrt{7})^2-(\sqrt{2})^2=7-2=5\). Hence the value is 5. The closest distractor 9 is wrong because it is the sum \(7+2\), i.e. sums of squares, not the product of conjugates. \(\sqrt{14}\) is incorrect since \(\sqrt{7}\cdot\sqrt{2}=\sqrt{14}\) holds for the product of the individual radicals, but not for the product of the summed expressions. \(9+2\sqrt{14}\) is the expansion of \((\sqrt{7}+\sqrt{2})^2\), not the product with its conjugate. Exam tip: spot conjugate pairs and apply \(x^2-y^2\) to avoid extra algebraic expansion.

(\frac{\sqrt{3}}{\sqrt{12}}=\sqrt{\frac{3}{12}}=\sqrt{\frac{1}{4}}=\frac{1}{2}). Simplify the ratio inside the roots.

Direct answer: option A, \(\sqrt2+\sqrt3>\sqrt5\). Both sides are positive, so squaring preserves the comparison. First, \((\sqrt2+\sqrt3)^2=2+3+2\sqrt6=5+2\sqrt6\). Since \(\sqrt6>0\), this is greater than 5, which is \((\sqrt5)^2\). Therefore the original left side is greater than \(\sqrt5\). Option A is correct. Option B reverses the proven inequality. Option C says equality, but the extra positive term \(2\sqrt6\) shows the squares are unequal. Option D is wrong because both quantities are real, positive, and directly comparable. A useful caution is that squaring an inequality is safe here because both sides are non-negative. Memory cue: square positive surd expressions, expand, and compare.

\(\sqrt{99}\) ≈ 9.949874... and \(\sqrt{100}=10\). 9.98 lies between them because \(9.98^2=99.6004\), which is between 99 and 100. A finite decimal like 9.98 is rational (e.g. \(\frac{998}{100}\)). Why other choices fail: 9.8 is below \(\sqrt{99}\) (\(9.8^2=96.04\)), 10.1 is above 10 (\(10.1^2=102.01\)), and \(\sqrt{101}\) is irrational and greater than 10. Exam tip: to test if a number x lies between \(\sqrt{a}\) and \(\sqrt{b}\), compare \(x^2\) with a and b — this avoids decimal rounding errors.

(\sqrt{28}=2\sqrt{7}) and (\sqrt{175}=5\sqrt{7}). Thus (4\sqrt{7}+4\sqrt{7}-5\sqrt{7}=3\sqrt{7}).

Square the sum: \(x^2=(\sqrt{3}+\sqrt{2})^2=3+2+2\sqrt{6}=5+2\sqrt{6}\). Hence \(x^2-5=2\sqrt{6}\). Option C (\(4\sqrt{6}\)) is wrong because the middle term was doubled again; option D has the wrong sign; option B (0) is inconsistent since the radicals do not cancel. Exam tip: when expanding \((a+b)^2\), remember the middle term is \(2ab\).

To rationalize a denominator containing a square root, multiply the numerator and denominator by the conjugate of the denominator. The conjugate changes the sign between the terms, and the product then uses the difference-of-squares identity. This removes the radical from the denominator without changing the value of the fraction, because the multiplying factor is effectively 1.

The conjugate of \(5-\sqrt{6}\) is \(5+\sqrt{6}\). Hence \(\frac{1}{5-\sqrt{6}}\cdot\frac{5+\sqrt{6}}{5+\sqrt{6}}=\frac{5+\sqrt{6}}{25-6}=\frac{5+\sqrt{6}}{19}\). Therefore option A is correct. The denominator 31 results from an incorrect sum of squares, while option D omits the required denominator.

The terms become (\sqrt{5}+2\sqrt{5}+3\sqrt{5}+4\sqrt{5}). The total is (10\sqrt{5}), so check the options carefully.

Square \(\sqrt{6}+2\): \((\sqrt{6}+2)^2=6+4+4\sqrt{6}=10+4\sqrt{6}\). Since \(\sqrt{96}=4\sqrt{6}\), we get \((\sqrt{6}+2)^2=10+\sqrt{96}\), so \(\sqrt{10+\sqrt{96}}=\sqrt{6}+2\). Option B gives the wrong sign in the cross-term (\((2-\sqrt{6})^2=10-\sqrt{96}\)). Option C squares to 18, not the given expression. Option D is not equal to the required simplified value. Exam tip: look for representation \(\sqrt{m}+\sqrt{n}\) so that \((\sqrt{m}+\sqrt{n})^2=m+n+2\sqrt{mn}\) matches the given form; match the cross-term to identify m and n.

Option A is correct. Compute \(x^2=(\sqrt{7}+\sqrt{3})^2=7+3+2\sqrt{7}\sqrt{3}=10+2\sqrt{21}\), so \(x^2-10=2\sqrt{21}\). Option B (0) is wrong — that would follow only if the cross term \(2\sqrt{21}\) were omitted. Option D (\(\sqrt{21}\)) is incorrect because the cross term equals \(2\sqrt{21}\), not its half. Exam tip: expand the square using \((a+b)^2=a^2+2ab+b^2\) and always include the cross term when surds are involved.

The zeroes are (x=\pm\sqrt{2}), and (\sqrt{2}) is irrational. In exams, simplify square-root zeroes before deciding the type.

For a quadratic with rational coefficients, (a-\sqrt{b}) accompanies (a+\sqrt{b}). Remember this as the conjugate-zero rule.

A linear polynomial has the form \(ax+b\) and its zero is \(-\tfrac{b}{a}\). For a polynomial \(x-\alpha\), the zero is \(\alpha\). Setting \(x-2\sqrt{3}=0\) (option A) gives \(x=2\sqrt{3}\), so A is correct. Option B gives \(x=-2\sqrt{3}\) (wrong sign). Option C yields \(x=\tfrac{\sqrt{3}}{2}\), not \(2\sqrt{3}\). Option D is quadratic (degree 2), not linear; its zeros are ±\(2\sqrt{3}\). Exam tip: For quick checks use \(x=-b/a\) for \(ax+b\) and verify the degree is 1.

The discriminant is (D=36-16=20), and (20) is not a perfect square. So the zeroes are real, distinct, and irrational.

The sum of the zeros is \\(\sqrt{7}+(-\sqrt{7})=0\\) and the product is \\(\sqrt{7}\times(-\sqrt{7})=-7\\). For a monic quadratic the form is \\(x^2-(\text{sum})x+(\text{product})\\), so the required polynomial is \\(x^2-7\\). Why other options fail: option B has constant +7 (product +7) giving imaginary roots; option C has sum \\(2\sqrt{7}\\) and product +7 (a double positive root), not \\(\\pm\sqrt{7}\\); option D matches the product -7 but its sum is \\(-2\sqrt{7}\\), not 0, so its roots are different. Exam tip: compute sum and product of given roots and substitute into \\(x^2-(\text{sum})x+(\text{product})\\).

Use the relation for a monic quadratic: if roots are r1 and r2, polynomial is \(x^2-(r1+r2)x+(r1r2)\). Here sum = \((2+\sqrt{3})+(2-\sqrt{3})=4\) and product = \((2+\sqrt{3})(2-\sqrt{3})=4-3=1\). Thus the polynomial is \(x^2-4x+1\). Option B is wrong due to the wrong sign on the linear term (+4x instead of -4x); options C and D have different sum/product values. Exam tip: compute sum and product first and then form \(x^2-(\text{sum})x+\text{product}\); verify by substituting one root.

The sum of the roots is \((2+\sqrt{3})+(2-\sqrt{3})=4\). For the quadratic \(x^2-kx+1\), by Vieta the sum of roots equals \(k\) (since coefficient of x is \(-k\)). Hence \(k=4\). As a consistency check, the product is \((2+\sqrt{3})(2-\sqrt{3})=1\), matching the constant term. Exam tip: for conjugate roots of the form \(a\pm\sqrt{b}\), the sum is \(2a\); use Vieta quickly to identify coefficients.

The quadratic factors as \(x^2-2\sqrt{5}x+5=(x-\sqrt{5})^2\). Hence the root \(\sqrt{5}\) has multiplicity two and both zeroes equal \(\sqrt{5}\). The closest distractor (option C) is wrong because the polynomial does not have two distinct roots; option D is wrong because the discriminant is zero, so a real repeated root exists. Exam tip: compute the discriminant \(b^2-4ac\); if it equals zero, the quadratic has a repeated real root.