Rationalizing and Radical Conjugates: Practice

A sign flip fails here: $(1 + \sqrt[3]{2})(1 - \sqrt[3]{2}) = 1 - \sqrt[3]{4}$, still irrational. Use the sum-of-cubes multiplier $x^2 - xy + y^2$ with $x = 1$ and $y = \sqrt[3]{2}$, which is $1 - \sqrt[3]{2} + \sqrt[3]{4}$.

$\left(1 + \sqrt[3]{2}\right)\left(1 - \sqrt[3]{2} + \sqrt[3]{4}\right) = 1^3 + \left(\sqrt[3]{2}\right)^3 = 1 + 2 = 3$

So the answer is $\frac{1 - \sqrt[3]{2} + \sqrt[3]{4}}{3}$. The minus sign in the middle of the multiplier is what makes every intermediate term cancel; using all plus signs would leave radicals behind.

Multiply top and bottom by $\sqrt{10} + 3$, which rationalizes the numerator.

$\sqrt{10} - 3 = \frac{(\sqrt{10} - 3)(\sqrt{10} + 3)}{\sqrt{10} + 3} = \frac{10 - 9}{\sqrt{10} + 3} = \frac{1}{\sqrt{10} + 3}$

Since the conjugate product is $1$, the numbers $\sqrt{10} - 3$ and $\sqrt{10} + 3$ are reciprocals. The value is about $0.1623$, close to $\frac{1}{6} \approx 0.1667$ but not equal to it.

Rationalize each one and compare canonical forms. Since $(2 + \sqrt{3})(2 - \sqrt{3}) = 1$, the first two are equal:

$\frac{1}{2 + \sqrt{3}} = \frac{2 - \sqrt{3}}{1} = 2 - \sqrt{3} \approx 0.268$

The fourth is the same number in disguise, because $8 + 4\sqrt{3} = 4(2 + \sqrt{3})$, so $\frac{4}{8 + 4\sqrt{3}} = \frac{1}{2 + \sqrt{3}} = 2 - \sqrt{3}$. The third, however, rationalizes to $2 + \sqrt{3} \approx 3.732$, the reciprocal of the others rather than the same number.

Rationalize each fraction. Both conjugate products equal $3 - 2 = 1$, so each fraction collapses to a single surd expression.

$\frac{1}{\sqrt{3} - \sqrt{2}} = \sqrt{3} + \sqrt{2}, \qquad \frac{1}{\sqrt{3} + \sqrt{2}} = \sqrt{3} - \sqrt{2}$

Adding them, the $\sqrt{2}$ terms cancel and the $\sqrt{3}$ terms double:

$(\sqrt{3} + \sqrt{2}) + (\sqrt{3} - \sqrt{2}) = 2\sqrt{3} \approx 3.464$

Rationalize each fraction first, exactly as before. Each conjugate product is $3 - 2 = 1$.

$\frac{1}{\sqrt{3} - \sqrt{2}} = \sqrt{3} + \sqrt{2}, \qquad \frac{1}{\sqrt{3} + \sqrt{2}} = \sqrt{3} - \sqrt{2}$

Now subtract, and this time the $\sqrt{3}$ terms cancel while the $\sqrt{2}$ terms double:

$(\sqrt{3} + \sqrt{2}) - (\sqrt{3} - \sqrt{2}) = 2\sqrt{2} \approx 2.828$

The answer is not $0$: the two fractions look symmetric but they are not equal, since $\sqrt{3} - \sqrt{2} \approx 0.318$ and $\sqrt{3} + \sqrt{2} \approx 3.146$.

Multiply top and bottom by the conjugate $5 + 2\sqrt{6}$. The denominator is $5^2 - 2^2(6) = 25 - 24 = 1$.

$\frac{1}{5 - 2\sqrt{6}} = \frac{5 + 2\sqrt{6}}{1} = 5 + 2\sqrt{6}$

By the uniqueness of the form $a + b\sqrt{6}$, we can read off $a = 5$ and $b = 2$, so $a + b = 7$. Check the size: $5 - 2\sqrt{6} \approx 0.101$ and $1 \div 0.101 \approx 9.899 \approx 5 + 2(2.449)$.

Multiply top and bottom by the conjugate $\sqrt{6} + \sqrt{2}$. The denominator becomes $6 - 2 = 4$, and the numerator distributes.

$\sqrt{2}\left(\sqrt{6} + \sqrt{2}\right) = \sqrt{12} + 2 = 2\sqrt{3} + 2$

Now every term shares a factor of $2$, so cancel it:

$\frac{2\sqrt{3} + 2}{4} = \frac{\sqrt{3} + 1}{2} \approx 1.366$

That matches $1.414 \div 1.035 \approx 1.366$.

Rationalize $x$ first. Since $(2 - \sqrt{3})(2 + \sqrt{3}) = 4 - 3 = 1$, the reciprocal is immediate.

$x = \frac{2 + \sqrt{3}}{1} = 2 + \sqrt{3}, \qquad \frac{1}{x} = 2 - \sqrt{3}$

Adding them, the surd parts are opposite and cancel:

$(2 + \sqrt{3}) + (2 - \sqrt{3}) = 4$

This is the conjugate pair doing its other job: a number and its conjugate always add to the rational number $2a$.

Write the denominator as a power of $5$: $\sqrt[3]{25} = 5^{2/3}$. The exponent needed to complete the cube is $\tfrac{1}{3}$, so multiply by $\sqrt[3]{5}$.

$\frac{1}{5^{2/3}}\cdot\frac{5^{1/3}}{5^{1/3}} = \frac{5^{1/3}}{5^{1}} = \frac{\sqrt[3]{5}}{5}$

Numerically $1 \div 2.924 \approx 0.342$, and $1.710 \div 5 \approx 0.342$. Multiplying by $\sqrt[3]{25}$ instead would give $\sqrt[3]{625}$ downstairs, which is still irrational.

Rationalize the numerator to turn a delicate subtraction into a division.

$\sqrt{65} - 8 = \frac{65 - 64}{\sqrt{65} + 8} = \frac{1}{\sqrt{65} + 8}$

Add $8$ throughout the given bounds to get $16 < \sqrt{65} + 8 < 16.1$. A bigger denominator makes a smaller fraction, so taking reciprocals reverses the inequality:

$\frac{1}{16.1} < \sqrt{65} - 8 < \frac{1}{16}$

That traps the answer between about $0.0621$ and $0.0625$, and the true value is $0.0623$.

Multiply top and bottom by the conjugate $1 + \sqrt{2}$. The denominator becomes $1 - 2 = -1$, and the numerator expands.

$(3 + \sqrt{2})(1 + \sqrt{2}) = 3 + 3\sqrt{2} + \sqrt{2} + 2 = 5 + 4\sqrt{2}$

Dividing by $-1$ flips every sign:

$\frac{5 + 4\sqrt{2}}{-1} = -5 - 4\sqrt{2}$

So $a = -5$ and $b = -4$, and the question asks for $b$. Check the size: $(3 + 1.414) \div (1 - 1.414) \approx -10.66$, matching $-5 - 4(1.414)$.

Suppose $p + q\sqrt{5} = p' + q'\sqrt{5}$ and collect the terms.

$(q - q')\sqrt{5} = p' - p$

If $q \ne q'$ we could divide by the nonzero rational $q - q'$ and conclude that $\sqrt{5}$ is a quotient of rationals, hence rational, which is false. So $q = q'$, and the equation then forces $p = p'$. The converse is obvious. That uniqueness is exactly what makes the rationalized form canonical, and it is why the other statements fail: $\pi$ has no such form, $\sqrt{5} = 0 + 1\sqrt{5}$ has only one, and sums stay in the family since $(p + p') + (q + q')\sqrt{5}$ is the same shape.