4.2 Radicals & Operations with Roots

Quick Answer: A radical is a root, which is a fractional exponent. On GMAT Quantitative Reasoning — 21 questions, 45 minutes, no calculator, Official Guide Quantitative Review 2026–2027 Math Review 3.2 — $\sqrt{x}$ means the principal (non-negative) square root. The identity $\sqrt{x^2} = |x|$ is the highest-yield trap in this family: it equals $-x$ whenever $x < 0$. Multiply and divide radicals by combining radicands; add them only when the simplified radicands and indices match. Clear a binomial radical denominator with its conjugate. Nested roots multiply indices. Even roots of negatives are undefined in the reals.

Principal Square Root and the Absolute-Value Identity

The radical symbol is never the negative root by itself

• $\sqrt{x}$ is defined as the unique non-negative number whose square is $x$ (for $x \ge 0$).
• The equation $t^2 = 49$ has two solutions, $t = 7$ and $t = -7$. The expression $\sqrt{49}$ has one value: $7$. The negative root is written $-\sqrt{49} = -7$.

$\sqrt{x^2} = |x|$

$$\sqrt{x^2} = |x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$$

• If $x = 5$, $\sqrt{5^2} = \sqrt{25} = 5 = |5|$.
• If $x = -5$, $\sqrt{(-5)^2} = \sqrt{25} = 5 = |-5| = -x$, not $-5$.
• If $x < 0$, $\sqrt{x^2} + x = |x| + x = -x + x = 0$, not $2x$.

The same absolute-value wrap applies to any even root: $\sqrt[4]{x^4} = |x|$. Odd roots do not wrap: $\sqrt[3]{x^3} = x$ for every real $x$, including negatives.

Simplifying Radicals by Perfect-Power Factors

To simplify a square root, factor out the largest perfect-square factor of the radicand ($4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, \ldots$) and use $\sqrt{AB} = \sqrt{A}\sqrt{B}$ for $A, B \ge 0$:

• $\sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2}$.
• $\sqrt{180} = \sqrt{36 \cdot 5} = 6\sqrt{5}$.
• $\sqrt{450} = \sqrt{225 \cdot 2} = 15\sqrt{2}$.
• For non-negative variables, $\sqrt{50 x^5 y^6} = \sqrt{25 \cdot 2 \cdot x^4 \cdot x \cdot y^6} = 5 x^2 y^3 \sqrt{2x}$.

Leaving a perfect square inside the radical is incomplete simplification, and GMAT answer forms are almost always fully simplified.

Operations with Radicals

Multiplication and division

For $a \ge 0$ and $b > 0$:

$$\sqrt{a}\sqrt{b} = \sqrt{ab}, \qquad \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$$

• $\sqrt{12}\sqrt{6} = \sqrt{72} = 6\sqrt{2}$.
• Decompose first: $\sqrt{12}\sqrt{6} = (2\sqrt{3})(\sqrt{2}\sqrt{3}) = 2 \cdot 3 \sqrt{2} = 6\sqrt{2}$.
• $\dfrac{\sqrt{150}}{\sqrt{6}} = \sqrt{25} = 5$.

These product and quotient rules fail in the reals if an even-root radicand is negative, so do not apply them blindly to signed expressions until you have checked the domain.

Addition and subtraction: like radicals only

$$k\sqrt{r} + m\sqrt{r} = (k+m)\sqrt{r}$$

Unlike radicals — different indices or different simplified radicands — do not combine. $\sqrt{2} + \sqrt{3}$ is already simplified; it is not $\sqrt{5}$.

Protocol:

1. Simplify every term completely.
2. Add coefficients of identical remaining radicals.
3. Leave unlike terms side by side.

Worked combination:

$$3\sqrt{8} + 5\sqrt{18} - \sqrt{50} + \sqrt{27}$$

• $3\sqrt{8} = 3 \cdot 2\sqrt{2} = 6\sqrt{2}$
• $5\sqrt{18} = 5 \cdot 3\sqrt{2} = 15\sqrt{2}$
• $\sqrt{50} = 5\sqrt{2}$
• $\sqrt{27} = 3\sqrt{3}$

Result: $(6 + 15 - 5)\sqrt{2} + 3\sqrt{3} = 16\sqrt{2} + 3\sqrt{3}$.

Rationalizing Denominators

GMAT choices almost never leave an even root in a denominator.

Monomial denominator

$$\frac{a}{\sqrt{b}} = \frac{a\sqrt{b}}{b} \quad (b > 0)$$

• $\dfrac{15}{\sqrt{5}} = 3\sqrt{5}$.
• $\dfrac{4}{3\sqrt{2}} = \dfrac{4\sqrt{2}}{6} = \dfrac{2\sqrt{2}}{3}$.

Binomial denominator: multiply by the conjugate

$$(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$$ $$(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b$$

Rationalize $\dfrac{6}{\sqrt{5} - \sqrt{2}}$:

$$\frac{6(\sqrt{5} + \sqrt{2})}{5 - 2} = 2(\sqrt{5} + \sqrt{2}) = 2\sqrt{5} + 2\sqrt{2}.$$

Rationalize $\dfrac{4}{3 - \sqrt{7}}$:

$$\frac{4(3 + \sqrt{7})}{9 - 7} = 2(3 + \sqrt{7}) = 6 + 2\sqrt{7}.$$

The conjugate of $a + \sqrt{b}$ is $a - \sqrt{b}$, not $-a + \sqrt{b}$. Using the wrong partner leaves a radical in the denominator.

Higher-Index Roots and Nested Radicals

Nested roots multiply indices:

$$\sqrt[m]{\sqrt[n]{x}} = \sqrt[mn]{x} = x^{1/(mn)}$$

• $\sqrt{\sqrt[3]{64}} = \sqrt[6]{64} = \sqrt[6]{2^6} = 2$.
• $\sqrt[3]{\sqrt{x}} = x^{1/6} = \sqrt[6]{x}$.
• $\sqrt{\sqrt{81}} = 81^{1/4} = 3$, matching $(\sqrt{81})^{1/2} = 9^{1/2} = 3$.

A nested product such as $\sqrt{a\sqrt{a}}$ is $a^{3/4}$ for $a \ge 0$, because $\sqrt{a \cdot a^{1/2}} = (a^{3/2})^{1/2} = a^{3/4}$. Convert nested radicals to exponents whenever the index arithmetic is clearer than the radical picture.

Cube-root arithmetic follows the same product rule with index $3$: $\sqrt[3]{16}\sqrt[3]{4} = \sqrt[3]{64} = 4$. Do not mix indices in a single radical symbol without converting to exponents first; $\sqrt{x}\sqrt[3]{x} = x^{1/2}x^{1/3} = x^{5/6} = \sqrt[6]{x^5}$.

Master Radical Operations Table