Operations with Radicals: Practice

Multiply each term of the first factor against each term of the second. Two of the four products lose their radicals because a radical meets its twin.

$3\sqrt2 \cdot \sqrt2 = 6, \quad 3\sqrt2 \cdot 2\sqrt7 = 6\sqrt{14}, \quad -\sqrt7 \cdot \sqrt2 = -\sqrt{14}, \quad -\sqrt7 \cdot 2\sqrt7 = -14$

Collect the rational terms and the like radicals separately.

$(6 - 14) + (6 - 1)\sqrt{14} = -8 + 5\sqrt{14}$

Numerically that is about $-8 + 18.71 = 10.71$, matching $1.597 \times 6.706 \approx 10.71$ from the original factors.

Both radicands are nonnegative, so the product rule lets you split off the perfect square in each.

$\sqrt{9x} = \sqrt9 \cdot \sqrt x = 3\sqrt x, \qquad \sqrt{25x} = \sqrt{25} \cdot \sqrt x = 5\sqrt x$

The two terms are like radicals with the shared unit $\sqrt x$, so the coefficients add.

$3\sqrt x + 5\sqrt x = 8\sqrt x$

Check with $x = 4$: the original is $\sqrt{36} + \sqrt{100} = 6 + 10 = 16$, and $8\sqrt4 = 16$.

Do the division first, using the quotient rule with nonnegative radicands.

$\frac{\sqrt{48}}{\sqrt3} = \sqrt{\frac{48}{3}} = \sqrt{16} = 4$

Now simplify the remaining radical.

$\sqrt{27} = 3\sqrt3$

The sum is $4 + 3\sqrt3$. The $4$ is rational and $3\sqrt3$ is not, so they are unlike terms and cannot be merged into $7\sqrt3$.

The product rule $\sqrt a \cdot \sqrt b = \sqrt{ab}$ carries the hypothesis $a \ge 0$ and $b \ge 0$, and both radicands here are negative, so the rule simply does not apply.

$\sqrt{-4}\cdot\sqrt{-9} = (2i)(3i) = 6i^2 = -6 \ne 6 = \sqrt{36}$

The two answers differ in sign, which is exactly why the hypothesis is stated. The other explanations are wrong on their own terms: like radicals govern addition, not multiplication, and the radical sign always denotes the single principal root, so $\sqrt{36}$ is $6$ and nothing else.

A cube root of a negative number is a real number, because an odd power of a negative is negative. Evaluate each root from its definition.

$\sqrt[3]{-8} = -2 \ \text{ since } (-2)^3 = -8, \qquad \sqrt[3]{-27} = -3 \ \text{ since } (-3)^3 = -27$

Now multiply the two real numbers.

$(-2)(-3) = 6$

Unlike the square-root case, the product rule survives here: $\sqrt[3]{(-8)(-27)} = \sqrt[3]{216} = 6$ agrees. Odd indices accept negative radicands; even indices do not.

Handle each product first with the product rule; both come out as perfect squares.

$\sqrt{50}\cdot\sqrt2 = \sqrt{100} = 10, \qquad \sqrt{18}\cdot\sqrt2 = \sqrt{36} = 6$

The radicals are gone before any subtraction happens, so finish in whole numbers.

$10 - 6 = 4$

Factoring the $\sqrt2$ out first is an equally good route: $\sqrt2(\sqrt{50} - \sqrt{18}) = \sqrt2(5\sqrt2 - 3\sqrt2) = \sqrt2 \cdot 2\sqrt2 = 4$.

Build the cube from the square rather than trying to do it in one step. First square the binomial.

$(1 + \sqrt2)^2 = 1 + 2\sqrt2 + 2 = 3 + 2\sqrt2$

Now multiply that by one more factor of $(1 + \sqrt2)$, expanding term by term.

$(3 + 2\sqrt2)(1 + \sqrt2) = 3 + 3\sqrt2 + 2\sqrt2 + 2 \cdot 2 = 7 + 5\sqrt2$

As a check, $(1 + 1.414)^3 \approx 2.414^3 \approx 14.07$, and $7 + 5\sqrt2 \approx 14.07$.

Both radicands are nonnegative because $x \ge 0$, so the product rule applies.

$\sqrt{8x} \cdot \sqrt{2x} = \sqrt{8x \cdot 2x} = \sqrt{16x^2}$

Now simplify, remembering that the principal square root of $x^2$ is $|x|$.

$\sqrt{16x^2} = 4|x| = 4x \quad \text{(since } x \ge 0\text{)}$

Without the hypothesis $x \ge 0$ the answer would have to be written $4|x|$. Check with $x = 9$: $\sqrt{72}\cdot\sqrt{18} = \sqrt{1296} = 36 = 4 \cdot 9$.

Simplify each radical, then multiply through by the outside coefficients.

$3\sqrt{45} = 3 \cdot 3\sqrt5 = 9\sqrt5, \qquad 2\sqrt{80} = 2 \cdot 4\sqrt5 = 8\sqrt5, \qquad \sqrt{125} = 5\sqrt5$

All three are multiples of $\sqrt5$, so combine the coefficients with their signs.

$9\sqrt5 - 8\sqrt5 + 5\sqrt5 = (9 - 8 + 5)\sqrt5 = 6\sqrt5$

Every wrong answer here is a bookkeeping slip: losing the minus sign gives $9 + 8 + 5 = 22$, dropping the middle term gives $9 + 5 = 14$, and forgetting that $\sqrt{125} = 5\sqrt5$ gives $9 - 8 + 1 = 2$. Keep every sign and coefficient attached through each step.

This is a difference times the matching sum, so square each term and subtract, squaring the coefficient along with the radical.

$5^2 - (2\sqrt6)^2 = 25 - 4 \cdot 6 = 25 - 24 = 1$

The cross terms $+10\sqrt6$ and $-10\sqrt6$ cancel, so no radical survives. Notice how close the two factors are to reciprocals: $5 - 2\sqrt6 \approx 0.101$ and $5 + 2\sqrt6 \approx 9.899$, and their product really is $1$.

Only the conjugate pair has cross terms that cancel, and cancelling them is what removes the radical.

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

Each of the others keeps an irrational $\sqrt{15}$ or a bare radical: $(\sqrt5 - \sqrt3)^2 = 8 - 2\sqrt{15}$ and $\sqrt5 \cdot \sqrt3 = \sqrt{15}$, and $15$ is not a perfect square, so $\sqrt{15}$ is irrational. For the last one, if $\sqrt5 + \sqrt3$ were rational then its square $8 + 2\sqrt{15}$ would be rational too, forcing $\sqrt{15}$ to be rational, which it is not.

The indices $4$ and $2$ differ, so convert to rational exponents over the common base $3$, using $9 = 3^2$.

$\sqrt[4]{9} = (3^2)^{1/4} = 3^{2/4} = 3^{1/2}$

The fourth root of $9$ is just $\sqrt3$ in disguise, so the product is a radical times itself.

$3^{1/2} \cdot 3^{1/2} = 3^{1} = 3$

Reducing $2/4$ to $1/2$ is the whole trick: always simplify the fractional exponent before multiplying.