Operations with Radicals: Additional Practice (Set 2)

Simplify all three radicals before comparing them, since each hides a square factor.

$\sqrt{20} = 2\sqrt5, \qquad \sqrt{45} = 3\sqrt5, \qquad \sqrt{80} = 4\sqrt5$

Every term is a multiple of $\sqrt5$, so collect the coefficients.

$2\sqrt5 + 3\sqrt5 - 4\sqrt5 = (2 + 3 - 4)\sqrt5 = \sqrt5$

The coefficient came out $1$, so a single $\sqrt5$ is left, about $2.24$.

Pull the largest square factor out of each radicand before comparing them.

$\sqrt{300} = 10\sqrt3, \qquad \sqrt{108} = \sqrt{36 \cdot 3} = 6\sqrt3, \qquad \sqrt{27} = 3\sqrt3$

All three share the unit $\sqrt3$, so combine the coefficients with their signs attached.

$10\sqrt3 - 6\sqrt3 + 3\sqrt3 = (10 - 6 + 3)\sqrt3 = 7\sqrt3$

Operating on the radicands instead gives $\sqrt{219} \approx 14.8$, while the true value is about $12.1$.

The indices $3$ and $4$ differ, so the product rule cannot be used. Convert to rational exponents.

$\sqrt[3]{2} \cdot \sqrt[4]{2} = 2^{1/3} \cdot 2^{1/4} = 2^{4/12 + 3/12} = 2^{7/12}$

The common denominator $12$ is the common index the radicals lacked, so translate back.

$2^{7/12} = \sqrt[12]{2^7} = \sqrt[12]{128}$

Adding the indices to get $\sqrt[7]{2}$ or multiplying them to get $\sqrt[12]{2}$ both ignore the numerators: it is the exponents that add.

The indices differ, so switch to rational exponents and add them over the common denominator $6$.

$5^{1/3} \cdot 5^{1/6} = 5^{2/6 + 1/6} = 5^{3/6} = 5^{1/2}$

The fraction $3/6$ reduces to $1/2$, so the answer is an ordinary square root.

$5^{1/2} = \sqrt5$

Adding the indices ($9$) or multiplying them ($18$) both ignore what the exponents actually do.

The bases differ and the indices differ, so rewrite both factors over the common index $6$, which is the least common multiple of $2$ and $3$.

$\sqrt2 = 2^{1/2} = 2^{3/6} = \sqrt[6]{2^3} = \sqrt[6]{8}, \qquad \sqrt[3]{3} = 3^{1/3} = 3^{2/6} = \sqrt[6]{3^2} = \sqrt[6]{9}$

Now the indices agree, so the product rule finally applies.

$\sqrt[6]{8} \cdot \sqrt[6]{9} = \sqrt[6]{72}$

Swapping the two exponents by mistake gives $\sqrt[6]{4 \cdot 27} = \sqrt[6]{108}$, which is about $2.18$ rather than the true $2.04$.

One factor is a square root and the other a fourth root, so the product rule cannot be used until the indices agree. Rational exponents settle it directly.

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

An exponent of $3/4$ is a fourth root of a cube.

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

The value is about $1.68$, which is indeed $1.414 \times 1.189$.

Squaring a sum produces three terms: the two squares and twice the cross product.

$(\sqrt3 + \sqrt5)^2 = (\sqrt3)^2 + 2\sqrt3\sqrt5 + (\sqrt5)^2 = 3 + 2\sqrt{15} + 5 = 8 + 2\sqrt{15}$

Answering $8$ squares term by term and discards the cross term. That cross term is exactly the $2\sqrt{ab}$ that makes $\sqrt a + \sqrt b = \sqrt{a+b}$ false, and it is worth about $7.75$ here, so it is not a rounding detail.

Both radicands are nonnegative, so split off the perfect square hiding in each coefficient.

$\sqrt{50x} = \sqrt{25 \cdot 2x} = 5\sqrt{2x}, \qquad \sqrt{18x} = \sqrt{9 \cdot 2x} = 3\sqrt{2x}$

The shared unit is $\sqrt{2x}$, so subtract the coefficients.

$5\sqrt{2x} - 3\sqrt{2x} = 2\sqrt{2x}$

Subtracting the radicands would give $\sqrt{32x}$: at $x = 2$ that is $8$, while the true value is $\sqrt{100} - \sqrt{36} = 4$.

The indices match, so the quotient rule applies to cube roots just as it does to square roots.

$\frac{\sqrt[3]{54}}{\sqrt[3]{2}} = \sqrt[3]{\frac{54}{2}} = \sqrt[3]{27} = 3$

The quotient of the radicands turned out to be a perfect cube, so the radical disappears. Subtracting the radicands to get $\sqrt[3]{52}$ is not a rule that exists.

Use the quotient rule on the fraction, since both radicands are nonnegative and the denominator is not zero.

$\frac{\sqrt{162}}{\sqrt2} = \sqrt{\frac{162}{2}} = \sqrt{81} = 9$

The second radical is a perfect square as well, so $\sqrt{49} = 7$ and both radicals disappear before the subtraction.

$9 - 7 = 2$

For cube roots, look for perfect cube factors rather than squares.

$\sqrt[3]{16} = \sqrt[3]{8 \cdot 2} = 2\sqrt[3]{2}, \qquad \sqrt[3]{54} = \sqrt[3]{27 \cdot 2} = 3\sqrt[3]{2}$

All three terms are now multiples of the same unit $\sqrt[3]{2}$ (index $3$, radicand $2$), so the coefficients combine.

$2\sqrt[3]{2} + 3\sqrt[3]{2} - \sqrt[3]{2} = (2 + 3 - 1)\sqrt[3]{2} = 4\sqrt[3]{2}$

The index must be carried through: $4\sqrt2 \approx 5.66$ is a different number from $4\sqrt[3]{2} \approx 5.04$.

The indices differ, so the quotient rule cannot be applied directly. Rewrite both roots as powers of $3$.

$\frac{\sqrt3}{\sqrt[3]{3}} = \frac{3^{1/2}}{3^{1/3}} = 3^{1/2 - 1/3}$

Dividing powers of the same base subtracts the exponents, so find the common denominator.

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

The result is about $1.20$, matching $1.732 / 1.442$.