Radicals

An entire radical has everything under the root sign, like $\sqrt{72}$. A mixed radical has a coefficient out front, like $6\sqrt{2}$. To simplify an entire radical to mixed form, find the largest perfect-square factor of the radicand, split it off, and take its root. This relies on the rule $\sqrt{ab}=\sqrt{a}\cdot \sqrt{b}$ for $a,b\ge 0$, which is true because squaring either side gives $ab$. It helps to know the perfect squares $4,9,16,25,36,49,64,81,100,121,144\text{.}$ For a large radicand, use prime factors instead and pull out each pair: $180=2\cdot 2\cdot 3\cdot 3\cdot 5$, and each pair leaves the root as one copy, so $\sqrt{180}=2\cdot 3\sqrt{5}=6\sqrt{5}$. A radical is fully simplified when the number left inside has no perfect-square factor other than $1$.

The process also runs backwards. To change a mixed radical to an entire radical, square the coefficient and multiply it inside: $3\sqrt{5}=\sqrt{9}\cdot \sqrt{5}=\sqrt{45}$. This is the easiest way to compare radicals without a calculator. Which is larger, $3\sqrt{5}$ or $4\sqrt{3}$? Convert both: $\sqrt{45}$ and $\sqrt{48}$, so $4\sqrt{3}$ is larger. One warning: the product rule has no sum version. $\sqrt{9+16}=\sqrt{25}=5$, but $\sqrt{9}+\sqrt{16}=3+4=7$, so $\sqrt{a+b}\ne \sqrt{a}+\sqrt{b}$.

Radicals combine like variables in algebra — only like radicals (same number under the root) can be added or subtracted, by combining their coefficients. Just as $5x+6x=11x$, we get $5\sqrt{2}+6\sqrt{2}=11\sqrt{2}$, because $\sqrt{2}$ is being counted like an object. But $\sqrt{2}+\sqrt{3}$ cannot be combined, any more than $x+y$ can. Radicals that don't look alike at first often become alike once simplified: $\sqrt{12}+\sqrt{27}=2\sqrt{3}+3\sqrt{3}=5\sqrt{3}$. So the order is always: simplify every radical, then group like radicals, then add their coefficients.

Unlike addition, multiplication works on any radicals of the same index, like or not. Multiply coefficients with coefficients and radicands with radicands, then simplify: $(2\sqrt{3})(5\sqrt{6})=10\sqrt{18}=10(3\sqrt{2})=30\sqrt{2}\text{.}$ Distribution and binomial products work exactly as in algebra, with $\sqrt{a}\cdot \sqrt{a}=a$: $\sqrt{3}(2+\sqrt{6})=2\sqrt{3}+\sqrt{18}=2\sqrt{3}+3\sqrt{2}$, and $(2+\sqrt{3})(4-\sqrt{3})=8-2\sqrt{3}+4\sqrt{3}-3=5+2\sqrt{3}\text{.}$ Squaring a binomial gives a middle term, so ${(1+\sqrt{2})}^2=1+2\sqrt{2}+2=3+2\sqrt{2}$, not $1+2$.

1.4 Radical Expressions (mcr3u)

A fraction is considered "not fully simplified" if it has a radical in the denominator. To fix a single-term radical denominator, multiply top and bottom by that same radical. When the denominator is a binomial with a radical (like $3+\sqrt{2}$), multiply top and bottom by its conjugate ($3-\sqrt{2}$) — the same two terms with the middle sign flipped — because $(A+B)(A-B)=A^2-B^2$ eliminates the radical entirely. The middle terms $-AB$ and $+AB$ cancel, and squaring the radical term leaves a whole number.

Why is multiplying top and bottom allowed? Because $\frac{\sqrt{7}}{\sqrt{7}}=1$ and $\frac{3-\sqrt{2}}{3-\sqrt{2}}=1$, so the value of the fraction never changes, only its appearance. For a single radical: $\frac{3}{\sqrt{7}}\cdot \frac{\sqrt{7}}{\sqrt{7}}=\frac{3\sqrt{7}}{7}$. With a coefficient, you only need the radical part: $\frac{5}{2\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{5\sqrt{3}}{6}$. The same skills tidy up quadratic-formula answers. For $x^2-4x-1=0$ the formula gives $\frac{4\pm \sqrt{20}}{2}$; simplify $\sqrt{20}=2\sqrt{5}$, then divide every term of the numerator by $2$ to get $2\pm \sqrt{5}$. Dividing only the $4$ (to get $2\pm \sqrt{20}$) is a common slip. Finally, check any simplified radical with a calculator: both forms must give the same decimal, for example $\frac{5\sqrt{3}}{6}\approx 1.443$ and $\frac{5}{2\sqrt{3}}\approx 1.443$.

Rationalize The Denominator