Operations with Radicals: Additional Practice (Set 1)

A difference times the matching sum is a difference of squares, so the cross terms $+2\sqrt6$ and $-2\sqrt6$ cancel.

$(\sqrt6 - 2)(\sqrt6 + 2) = (\sqrt6)^2 - 2^2 = 6 - 4 = 2$

The radical is gone. Adding the two squares instead of subtracting would give $10$.

Do not expand the squares. Regroup the four factors into two conjugate pairs first, which is legal because multiplication is commutative.

$(\sqrt5 + 2)^2(\sqrt5 - 2)^2 = \big[(\sqrt5 + 2)(\sqrt5 - 2)\big]^2$

The inner product is a difference of squares, and it collapses to a single number.

$\big[(\sqrt5)^2 - 2^2\big]^2 = (5 - 4)^2 = 1^2 = 1$

Expanding each square separately gives $(9 + 4\sqrt5)(9 - 4\sqrt5) = 81 - 80 = 1$ as well, but the regrouped route needs no radicals at all.

Expand each square, keeping the cross terms.

$(\sqrt7 + \sqrt3)^2 = 7 + 2\sqrt{21} + 3 = 10 + 2\sqrt{21}, \qquad (\sqrt7 - \sqrt3)^2 = 7 - 2\sqrt{21} + 3 = 10 - 2\sqrt{21}$

The two cross terms are negatives of each other, so adding the squares wipes them out.

$(10 + 2\sqrt{21}) + (10 - 2\sqrt{21}) = 20$

Squaring a sum keeps a radical and squaring a difference keeps a radical, yet their sum is a plain integer.

The product rule is stated for nonnegative radicands, and the reason shows up the moment a radicand goes negative.

$a = -4, \ b = -9: \quad \sqrt{-4}\cdot\sqrt{-9} = (2i)(3i) = -6, \quad \text{but} \quad \sqrt{(-4)(-9)} = \sqrt{36} = 6$

That single counterexample kills three of the four options at once: $a = -4$ and $b = -9$ are real, they have the same sign, and their product $ab = 36$ is nonnegative, yet the rule fails. Only $a \ge 0$ and $b \ge 0$ survives.

All three radicands are nonnegative, so the product rule extends to three factors: gather everything under one radical.

$\sqrt{12} \cdot \sqrt{18} \cdot \sqrt6 = \sqrt{12 \cdot 18 \cdot 6} = \sqrt{1296}$

Now recognise the perfect square, since $36^2 = 1296$.

$\sqrt{1296} = 36$

The answer $216$ is just $12 \cdot 18$, which drops the third factor $\sqrt6$ and never takes the root at all.

Distribute $\sqrt5$ across both terms, applying the product rule to each.

$\sqrt5 \cdot \sqrt{15} = \sqrt{75} = 5\sqrt3, \qquad \sqrt5 \cdot \sqrt{45} = \sqrt{225} = 15$

Keep the minus sign in front of the second product.

$5\sqrt3 - 15$

The value is about $8.66 - 15 = -6.34$, which is negative because $\sqrt{45} > \sqrt{15}$ made the bracket negative to begin with. Reversing the order to $15 - 5\sqrt3$ flips the sign and is a different number.

Square each term (coefficient included) and subtract twice the cross product.

$(3\sqrt2)^2 = 9 \cdot 2 = 18, \qquad (2\sqrt3)^2 = 4 \cdot 3 = 12, \qquad 2(3\sqrt2)(2\sqrt3) = 12\sqrt6$

Assemble the three pieces in the pattern $(u - v)^2 = u^2 - 2uv + v^2$.

$18 - 12\sqrt6 + 12 = 30 - 12\sqrt6$

The answer is about $0.61$, a small positive number, because $3\sqrt2 \approx 4.243$ and $2\sqrt3 \approx 3.464$ are close together. Answering $6$ treats $(u - v)^2$ as $u^2 - v^2$, giving $18 - 12$; dropping only the cross term would leave $18 + 12 = 30$.

The indices match, so the product rule applies: multiply the radicands under a single cube root.

$\sqrt[3]{4} \cdot \sqrt[3]{16} = \sqrt[3]{64}$

And $64$ is a perfect cube, since $4^3 = 64$.

$\sqrt[3]{64} = 4$

Neither factor was a nice number on its own, but their product was, which is why you simplify only after multiplying.

Simplify the left side first, since the two radicals turn out to be like radicals.

$\sqrt{20} + \sqrt{45} = 2\sqrt5 + 3\sqrt5 = 5\sqrt5$

Now push the coefficient back inside the radical, using $5 = \sqrt{25}$.

$5\sqrt5 = \sqrt{25}\cdot\sqrt5 = \sqrt{125}$

So $k = 125$, not $20 + 45 = 65$. The sum of these two radicals is a single square root, just not the one the false addition rule predicts.

Take the conjugate pair first; it collapses to a single number and makes the rest trivial.

$(2 + \sqrt3)(2 - \sqrt3) = 2^2 - (\sqrt3)^2 = 4 - 3 = 1$

The whole product is therefore $1$ times the remaining factor.

$1 \cdot (7 + 4\sqrt3) = 7 + 4\sqrt3$

Choosing the order of multiplication well is the entire trick here: expanding left to right would work but would cost three times the algebra.

This is a sum times the matching difference, so the cross terms $-3\sqrt x$ and $+3\sqrt x$ cancel.

$(\sqrt x + 3)(\sqrt x - 3) = (\sqrt x)^2 - 3^2 = x - 9$

The step $(\sqrt x)^2 = x$ needs $x \ge 0$, which is given. Check with $x = 16$: the original is $(4 + 3)(4 - 3) = 7$, and $16 - 9 = 7$. The choice $x - 6\sqrt x + 9$ is the expansion of $(\sqrt x - 3)^2$, where the cross terms reinforce instead of cancelling.

Three different indices, so the product rule is useless here. Write every factor as a power of $2$.

$\sqrt2 = 2^{1/2}, \qquad \sqrt[3]{4} = (2^2)^{1/3} = 2^{2/3}, \qquad \sqrt[6]{2} = 2^{1/6}$

Multiplying powers of one base means adding the exponents, over the common denominator $6$.

$2^{3/6} \cdot 2^{4/6} \cdot 2^{1/6} = 2^{8/6} = 2^{4/3}$

Since $4/3 = 1 + 1/3$, one whole factor of $2$ comes out.

$2^{4/3} = 2^1 \cdot 2^{1/3} = 2\sqrt[3]{2}$