Rationalizing and Radical Conjugates: Practice

A cube root needs the factor that completes the cube, not a sign flip. Since $\sqrt[3]{3} = 3^{1/3}$, the missing exponent is $\tfrac{2}{3}$, and $3^{2/3} = \sqrt[3]{9}$.

$\frac{1}{\sqrt[3]{3}}\cdot\frac{\sqrt[3]{9}}{\sqrt[3]{9}} = \frac{\sqrt[3]{9}}{\sqrt[3]{27}} = \frac{\sqrt[3]{9}}{3}$

Multiplying by $\sqrt[3]{3}$ instead would leave $\sqrt[3]{9}$ downstairs, which is still irrational.

Multiply top and bottom by the conjugate $3 - \sqrt{5}$. The denominator becomes $3^2 - (\sqrt{5})^2 = 9 - 5 = 4$.

$\frac{8}{3 + \sqrt{5}} = \frac{8(3 - \sqrt{5})}{4} = 2(3 - \sqrt{5}) = 6 - 2\sqrt{5}$

The $8$ and the $4$ reduce to a factor of $2$, so distribute it across both terms. Numerically $3 + \sqrt{5} \approx 5.236$ and $8 \div 5.236 \approx 1.528$, matching $6 - 2(2.236) \approx 1.528$.

Write the denominator as a power of $2$ and read off what is missing: $\sqrt[3]{4} = 2^{2/3}$, so the factor that completes the cube is $2^{1/3} = \sqrt[3]{2}$.

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

The mechanical rule "multiply by $\sqrt[3]{a^2}$" would use $\sqrt[3]{16}$ here and still work, but it drags in bigger numbers you then have to reduce. Checking the size: $1 \div 1.587 \approx 0.630$, and $1.260 \div 2 \approx 0.630$.

Multiply top and bottom by the conjugate $\sqrt{3} - 1$. The denominator becomes $3 - 1 = 2$, and the numerator distributes.

$\sqrt{3}(\sqrt{3} - 1) = 3 - \sqrt{3}$

So the answer is $\frac{3 - \sqrt{3}}{2}$. As a check, $\sqrt{3} \div (\sqrt{3} + 1) \approx 1.732 \div 2.732 \approx 0.634$, and $(3 - 1.732) \div 2 \approx 0.634$.

Rationalize the numerator by multiplying top and bottom by $\sqrt{50} + 7$. The numerator becomes a difference of squares.

$\sqrt{50} - 7 = \frac{(\sqrt{50} - 7)(\sqrt{50} + 7)}{\sqrt{50} + 7} = \frac{50 - 49}{\sqrt{50} + 7} = \frac{1}{\sqrt{50} + 7}$

That rewrite replaces a delicate subtraction of two nearly equal numbers with an easy division. Numerically both sides are about $0.0711$.

Multiply top and bottom by the conjugate $\sqrt{2} + \sqrt{3}$. The denominator is $2 - 3 = -1$, a negative rational, which is entirely allowed.

$\frac{\sqrt{6}(\sqrt{2} + \sqrt{3})}{-1} = \frac{\sqrt{12} + \sqrt{18}}{-1} = \frac{2\sqrt{3} + 3\sqrt{2}}{-1} = -2\sqrt{3} - 3\sqrt{2}$

The sign is the whole question here. Since $\sqrt{2} < \sqrt{3}$, the original fraction is negative, so a positive answer cannot be right.

Split the fraction and clear each piece separately.

$\frac{\sqrt{10}}{\sqrt{2}} + \frac{2}{\sqrt{2}} = \sqrt{\frac{10}{2}} + \frac{2\sqrt{2}}{2} = \sqrt{5} + \sqrt{2}$

Both terms rationalize on their own, and the second one collapses because $\frac{2}{\sqrt{2}} = \sqrt{2}$. Numerically the original is $(3.162 + 2) \div 1.414 \approx 3.650$, matching $2.236 + 1.414$.

A sign flip is useless against a cube root, so use the difference-of-cubes multiplier $x^2 + xy + y^2$ with $x = \sqrt[3]{2}$ and $y = 1$.

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

The denominator is $1$, so the answer is just the multiplier itself. Check the size: $\sqrt[3]{2} - 1 \approx 0.260$, so the original is about $3.85$, and $1.587 + 1.260 + 1 \approx 3.85$.

The cross terms $-\sqrt{ab}$ and $+\sqrt{ab}$ cancel, so only the squared terms survive, and each of those is rational.

$(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = (\sqrt{a})^2 - (\sqrt{b})^2 = a - b$

The last option, $a - 2\sqrt{ab} + b$, is the square $(\sqrt{a} - \sqrt{b})^2$, which keeps its radical. Squaring and conjugating are different operations.

Simplify the radical first: $\sqrt{12} = 2\sqrt{3}$, which keeps the arithmetic small.

$\frac{3}{2\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{3}}{2\cdot 3} = \frac{\sqrt{3}}{2}$

The $3$ upstairs cancels the $3$ that the multiplication created downstairs. Numerically $3 \div 3.464 \approx 0.866$, and $1.732 \div 2 \approx 0.866$.

Multiply top and bottom by the conjugate $2 + \sqrt{3}$. The denominator becomes $4 - 3 = 1$, and the numerator becomes a square.

$(2 + \sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}$

So the value is $7 + 4\sqrt{3} \approx 13.93$, which matches $3.732 \div 0.268 \approx 13.93$. The answer is not $1$: a fraction whose top and bottom are conjugates is only $1$ when they are equal, which they are not.

Write the denominator as a power of $3$: $\sqrt[3]{9} = 3^{2/3}$. The exponent needed to complete the cube is $\tfrac{1}{3}$, so the multiplier is $\sqrt[3]{3}$, not $\sqrt[3]{81}$.

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

Rational exponents make the choice of multiplier obvious: you only ever need whatever gets the exponent up to a whole number. Numerically $1 \div 2.080 \approx 0.481$, and $1.442 \div 3 \approx 0.481$.