Operations with Radicals: Practice

All three radicands are different, so simplify each one before deciding anything. Pull out the largest square factor from each.

$\sqrt{75} = 5\sqrt3, \qquad \sqrt{48} = 4\sqrt3, \qquad \sqrt{12} = 2\sqrt3$

Every term is now a multiple of the same unit $\sqrt3$, so the distributive law collects all three at once.

$5\sqrt3 + 4\sqrt3 - 2\sqrt3 = (5 + 4 - 2)\sqrt3 = 7\sqrt3$

Combining the radicands instead would give $\sqrt{75 + 48 - 12} = \sqrt{111} \approx 10.54$, but the true value is about $12.12$.

Coefficients multiply with coefficients and radicands with radicands, since multiplication is commutative and associative.

$(3\sqrt2)(5\sqrt6) = 15\sqrt{12}$

The product hides a square factor that neither original radical had, so simplify it.

$15\sqrt{12} = 15\sqrt{4 \cdot 3} = 15 \cdot 2\sqrt3 = 30\sqrt3$

Multiply each term of the first factor against each term of the second.

$2 \cdot 3 = 6, \quad 2 \cdot (-\sqrt5) = -2\sqrt5, \quad \sqrt5 \cdot 3 = 3\sqrt5, \quad \sqrt5 \cdot (-\sqrt5) = -5$

Now collect the rational terms and the like radicals separately.

$(6 - 5) + (-2 + 3)\sqrt5 = 1 + \sqrt5$

The cross terms do not cancel here because the signs of the two factors are not a matching sum-and-difference pair.

The indices differ, so the product rule is unavailable. Write both factors as powers of $2$ and add the exponents.

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

Over the common denominator $6$ the exponents add to $3/6 + 4/6 = 7/6$.

$2^{1/2} \cdot 2^{2/3} = 2^{7/6} = 2^1 \cdot 2^{1/6} = 2\sqrt[6]{2}$

Simplify each radical by pulling out its largest square factor.

$\sqrt{98} = 7\sqrt2, \qquad \sqrt{32} = 4\sqrt2, \qquad \sqrt8 = 2\sqrt2$

All three share the unit $\sqrt2$, so the coefficients combine.

$7\sqrt2 - 4\sqrt2 + 2\sqrt2 = (7 - 4 + 2)\sqrt2 = 5\sqrt2$

Working on the radicands instead would give $\sqrt{98 - 32 + 8} = \sqrt{74} \approx 8.60$, not the true value of about $7.07$.

Squaring a product squares both factors, so square the coefficient and square the radical separately.

$(4\sqrt3)^2 = 4^2 \cdot (\sqrt3)^2 = 16 \cdot 3 = 48$

The common slip is squaring only the radical and leaving the $4$ alone, which gives $12$.

For $a \ge 0$ and $b > 0$ the quotient rule says $\sqrt a / \sqrt b = \sqrt{a/b}$, so divide the radicands under one radical.

$\frac{\sqrt{72}}{\sqrt2} = \sqrt{\frac{72}{2}} = \sqrt{36} = 6$

The quotient of two irrational numbers came out a whole number because $72/2$ is a perfect square. Subtracting the radicands to get $\sqrt{70}$ is not an operation that exists.

Simplify each radical first, then multiply through by the outside coefficient.

$2\sqrt{45} = 2 \cdot 3\sqrt5 = 6\sqrt5, \qquad 3\sqrt{20} = 3 \cdot 2\sqrt5 = 6\sqrt5$

The two terms turn out to be identical, so their difference is zero.

$6\sqrt5 - 6\sqrt5 = 0$

Unequal-looking radical expressions can be exactly equal once simplified, which is why simplifying comes before comparing.

Distribute $\sqrt5$ across the sum, using the product rule on each piece.

$\sqrt5 \cdot \sqrt{10} = \sqrt{50} = 5\sqrt2, \qquad \sqrt5 \cdot \sqrt{20} = \sqrt{100} = 10$

So the expansion is

$5\sqrt2 + 10.$

One product kept its radical and the other did not, and a radical term never combines with a rational one.

Simplify each radical, remembering to carry the coefficient $3$ through the first one.

$3\sqrt8 = 3 \cdot 2\sqrt2 = 6\sqrt2, \qquad \sqrt{50} = 5\sqrt2$

Both are multiples of $\sqrt2$, so add the coefficients.

$6\sqrt2 + 5\sqrt2 = 11\sqrt2$

Forgetting to multiply the $3$ by the $2$ that comes out of $\sqrt8$ is what produces the wrong answer $8\sqrt2$.

The indices match, so the product rule applies directly to cube roots: multiply the radicands under one radical.

$\sqrt[3]{2} \cdot \sqrt[3]{4} = \sqrt[3]{2 \cdot 4} = \sqrt[3]{8} = 2$

The answer is a whole number because $8$ is a perfect cube. Adding the radicands to get $\sqrt[3]{6}$ is the addition error transplanted to a product.

Both radicands are nonnegative, so combine them under one radical and then simplify.

$\sqrt6 \cdot \sqrt{15} = \sqrt{90}$

Now pull out the largest square factor of $90$.

$\sqrt{90} = \sqrt{9 \cdot 10} = 3\sqrt{10}$

Neither factor had a square inside it, but their product did, which is why you always simplify after multiplying.