Simplest form, and getting a root off the bottom

$\sqrt{72}$ and $6\sqrt{2}$ are the same number; one of them is an answer and the other is not.

There is more than one way to write most numbers, and mathematics has a habit of preferring one of them. For roots the preference is called simplest form, and it is worth seeing that the choice really is between two spellings of one number before learning how to move between them.

The two spellings agree to every digit but the last, and that last one is the machine rounding differently on two different routes to one number. The move that gets from one to the other is a single rule.

The product and quotient rules for roots

These are the exponent laws $(ab)^{1/n} = a^{1/n}b^{1/n}$ in radical clothing, so there is nothing new to believe. What matters is that there is no matching rule for a sum: $\sqrt{a+b}$ cannot be split, as $\sqrt{9+16} = 5$ against $3 + 4 = 7$ shows.

A square root is in simplest form when the radicand has no perfect-square factor bigger than 1. So the job is to find the largest perfect square hiding inside, and take it out.

Higher roots and letters

For a cube root, look for perfect cubes instead: $\sqrt[3]{54} = \sqrt[3]{27 \cdot 2} = 3\sqrt[3]{2}$. Taking out 9 would achieve nothing, because 9 is a square and not a cube. For an index of $n$, the factor you want is a perfect $n$th power.

Letters behave the same way, and the exponent form shows which powers come out. Under a square root, an even power leaves cleanly, because half of an even number is a whole number: $\sqrt{x^{6}} = x^{3}$. An odd power leaves one factor behind: $\sqrt{x^{7}} = \sqrt{x^{6} \cdot x} = x^{3}\sqrt{x}$.

Under a cube root the test is divisibility by 3 rather than by 2, since a cube root undoes a third power. Here is one with a number and a letter together, every step.

Simplify $\sqrt[3]{54x^{5}}$, taking $x$ to be positive.

1. Split the radicand into a number part and a letter part: $\sqrt[3]{54} \cdot \sqrt[3]{x^{5}}$, using the product rule for roots.
2. The largest perfect cube dividing 54 is 27, because $54 = 27 \cdot 2$ and the next cube up, 64, is bigger than 54.
3. So $\sqrt[3]{54} = \sqrt[3]{27}\cdot\sqrt[3]{2} = 3\sqrt[3]{2}$.
4. For the letter, take the largest multiple of 3 that is no bigger than 5, which is 3: $x^{5} = x^{3}\cdot x^{2}$.
5. So $\sqrt[3]{x^{5}} = x \cdot \sqrt[3]{x^{2}}$, since $\sqrt[3]{x^{3}} = x$.
6. Put the two halves together and gather what is left under one root: $3x\sqrt[3]{2x^{2}}$.
7. Check at $x = 2$: the original radicand is $54 \cdot 32 = 1728$, and $\sqrt[3]{1728} = 12$. The answer gives $3 \cdot 2 \cdot \sqrt[3]{2\cdot4} = 6\sqrt[3]{8} = 6 \cdot 2 = 12$.

Two roots can be added only when they are like radicals, meaning the same index and the same radicand, in the same way that $3z + 5z = 8z$ but $3z + 5w$ does not collapse. The trap is that two roots can look unlike and become like once they are simplified, so simplify first and only then look.

Take $\sqrt{18} + \sqrt{50}$. Neither is simplified, and the radicands 18 and 50 are different, so nothing can be done yet. Simplify each: $18 = 9 \cdot 2$ gives $3\sqrt{2}$, and $50 = 25 \cdot 2$ gives $5\sqrt{2}$. Now both are multiples of $\sqrt{2}$, so add the numbers in front: $3 + 5 = 8$, and the sum is $8\sqrt{2}$.

Subtraction is the same move. $\sqrt{98} - \sqrt{18}$ becomes $7\sqrt{2} - 3\sqrt{2}$, since $98 = 49\cdot2$ and $18 = 9\cdot2$, and that is $4\sqrt{2}$. What never happens is the radicands themselves combining: $\sqrt{98}-\sqrt{18}$ is not $\sqrt{80}$, because $\sqrt{80} \approx 8.94$ while $4\sqrt{2} \approx 5.66$. The symbol $\approx$ means "is approximately equal to", and it is used whenever a decimal has been rounded.

Rationalising a denominator

By convention an answer does not leave a root on the bottom of a fraction. Rationalising the denominator means rewriting the fraction so the bottom is free of roots, without changing its value.

The tool is multiplication by 1, written in a useful disguise. Multiplying top and bottom by the same non-zero quantity is multiplying by 1, so the value cannot change, only the appearance. For $\frac{6}{\sqrt{3}}$, multiply top and bottom by $\sqrt{3}$: the bottom becomes $\sqrt{3}\cdot\sqrt{3} = 3$, the top becomes $6\sqrt{3}$, and the whole thing is $\frac{6\sqrt{3}}{3} = 2\sqrt{3}$.

When the bottom is a sum such as $2 - \sqrt{3}$, multiplying by $\sqrt{3}$ fails, because the 2 picks up a root of its own. What works is the conjugate: the same two terms with the sign between them flipped. The conjugate of $2 - \sqrt{3}$ is $2 + \sqrt{3}$, and multiplying them uses the pattern $(a-b)(a+b) = a^{2} - b^{2}$, which squares both terms and so destroys the root.

A worked example, every step

Rationalise the denominator of $\dfrac{6}{\sqrt{5}+\sqrt{2}}$.

1. The bottom is a sum of two terms, so the conjugate is the same pair with the sign flipped: $\sqrt{5}-\sqrt{2}$.
2. Multiply top and bottom by it. This is multiplying by $\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}$, which is 1, so the value is untouched.
3. The bottom is $(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})$. By the pattern $(a+b)(a-b) = a^{2}-b^{2}$ with $a = \sqrt{5}$ and $b = \sqrt{2}$, that is $5 - 2 = 3$.
4. The top is $6(\sqrt{5}-\sqrt{2})$.
5. So the fraction is $\dfrac{6(\sqrt{5}-\sqrt{2})}{3}$, and $6 \div 3 = 2$.
6. The answer is $2(\sqrt{5}-\sqrt{2})$, or $2\sqrt{5}-2\sqrt{2}$. The snippet above checks it: both forms print as 1.6437088302533893.