Operations with Radicals: Practice

The principal square root of $6$ is the nonnegative number whose square is $6$, so squaring it returns the radicand.

$(\sqrt6)^2 = \sqrt6 \cdot \sqrt6 = \sqrt{36} = 6$

Squaring and taking the square root are inverse operations on nonnegative numbers, which is why the radical disappears.

Multiplication is commutative and associative, so regroup the coefficients together and the radicals together.

$(2\sqrt5)(3\sqrt2) = (2 \cdot 3)(\sqrt5 \cdot \sqrt2) = 6\sqrt{10}$

Coefficients multiply with coefficients and radicands with radicands. Since $10$ has no square factor, $6\sqrt{10}$ is in simplest form.

The radicands differ, so simplify before judging. Split off the largest square factor of $18$.

$\sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt2$

Now both terms are multiples of the same unit, so the distributive law collects them.

$3\sqrt2 + \sqrt2 = (3 + 1)\sqrt2 = 4\sqrt2$

A radical times itself is the radical squared, and squaring undoes the principal square root.

$\sqrt5 \cdot \sqrt5 = \sqrt{5 \cdot 5} = \sqrt{25} = 5$

Watch the difference between multiplying and adding: $\sqrt5 \cdot \sqrt5 = 5$, while $\sqrt5 + \sqrt5 = 2\sqrt5$.

Simplify each sum first, then check whether the two terms share an index and a radicand.

$\sqrt3 + \sqrt3 = 2\sqrt3, \qquad 2\sqrt5 + 7\sqrt5 = 9\sqrt5, \qquad \sqrt2 + \sqrt8 = \sqrt2 + 2\sqrt2 = 3\sqrt2$

The first three all collect. In $\sqrt6 + \sqrt{10}$ neither radicand has a square factor, so both are already in simplest form with different radicands. There is no common unit to factor out, so that sum stays as written.

A bare $\sqrt7$ is $1\sqrt7$, so this is a subtraction of like radicals with coefficients $8$ and $1$.

$8\sqrt7 - 1\sqrt7 = (8 - 1)\sqrt7 = 7\sqrt7$

The unit $\sqrt7$ survives untouched; only the count changes.

A radical distributes over a sum like any other factor, so multiply $\sqrt2$ against each term inside the parentheses.

$\sqrt2 \cdot \sqrt2 + \sqrt2 \cdot 3 = 2 + 3\sqrt2$

The first product loses its radical because $\sqrt2 \cdot \sqrt2 = 2$, but the second keeps it. A rational term and a radical term are unlike, so $2 + 3\sqrt2$ cannot be collected further.

Simplify the first radical by pulling out the largest square factor of $27$.

$\sqrt{27} = \sqrt{9 \cdot 3} = 3\sqrt3$

The two terms are now like radicals, so add the coefficients.

$3\sqrt3 + \sqrt3 = (3 + 1)\sqrt3 = 4\sqrt3$

Adding the radicands to get $\sqrt{30}$ would give about $5.48$, but the true sum is about $6.93$.

This is a radical sum times the matching difference. Expanding gives cross terms $-\sqrt5\sqrt3$ and $+\sqrt3\sqrt5$, which cancel.

$(\sqrt5 + \sqrt3)(\sqrt5 - \sqrt3) = (\sqrt5)^2 - (\sqrt3)^2 = 5 - 3 = 2$

Both radicals vanish and the product of two irrational numbers is the integer $2$.

Both radicands are perfect squares, so evaluate each root separately and then add.

$\sqrt9 + \sqrt{16} = 3 + 4 = 7$

This is the standard refutation of the false rule $\sqrt a + \sqrt b = \sqrt{a+b}$: that rule would predict $\sqrt{9 + 16} = \sqrt{25} = 5$, and $5 \ne 7$.

The terms are like radicals with equal coefficients, so the distributive law gives a coefficient of zero.

$6\sqrt2 - 6\sqrt2 = (6 - 6)\sqrt2 = 0 \cdot \sqrt2 = 0$

A quantity minus itself is $0$, radical or not. The unit $\sqrt2$ is irrelevant once the coefficient hits zero.

Neither radicand is a perfect square, but each hides one. Pull out the largest square factor from each.

$\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt2, \qquad \sqrt8 = \sqrt{4 \cdot 2} = 2\sqrt2$

The two are like radicals once simplified, so subtract the coefficients.

$5\sqrt2 - 2\sqrt2 = 3\sqrt2$

Subtracting the radicands to get $\sqrt{42} \approx 6.48$ misses the true value of about $4.24$.