4.3 Exponent & Root Inequalities and Traps

Quick Answer: Exponents and roots do not keep one ordering on the whole real line. For $x > 1$, $\sqrt{x} < x < x^2 < x^3$. For $0 < x < 1$ the chain reverses: $x^3 < x^2 < x < \sqrt{x} < 1$. Negative bases split by parity: even powers are positive, odd powers stay negative, and even roots of negatives are undefined in the reals. Never write $\sqrt{a^2 + b^2} = a + b$ or $(ab)^n = a^n + b^n$. For $b^{f(x)} > b^{g(x)}$, keep the exponent inequality if $b > 1$ and reverse it if $0 < b < 1$. GMAT Quantitative Reasoning (21 questions, 45 minutes, no calculator; OG Quantitative Review 2026–2027 Math Review 3.2) tests these comparisons with exact arithmetic, not calculator graphs.

Four Intervals, Four Orderings

Mark $-1$, $0$, and $1$ on a number line before you compare $x$, $x^2$, $x^3$, and $\sqrt{x}$. Those three cut-points split the line into the four intervals GMAT writers actually use.

Interval 1: $x > 1$ (powers grow)

Multiplying by a number larger than $1$ increases magnitude:

$$\sqrt[3]{x} < \sqrt{x} < x < x^2 < x^3$$

Test $x = 4$: $\sqrt{4} = 2$, $4^2 = 16$, $4^3 = 64$, so $2 < 4 < 16 < 64$. Higher integer exponents make larger values; roots pull the value back toward $1$ from above.

Interval 2: $0 < x < 1$ (the reversal trap)

Multiplying by a proper fraction pulls the value toward $0$; taking a root pulls it toward $1$:

$$0 < x^3 < x^2 < x < \sqrt{x} < \sqrt[3]{x} < 1$$

Test $x = \dfrac{1}{4}$:

• $x^3 = \dfrac{1}{64} = 0.015625$
• $x^2 = \dfrac{1}{16} = 0.0625$
• $x = 0.25$
• $\sqrt{x} = 0.5$

So $\dfrac{1}{64} < \dfrac{1}{16} < \dfrac{1}{4} < \dfrac{1}{2} < 1$. Squaring a number in $(0,1)$ makes it smaller; taking a square root makes it larger.

The same reversal governs negative exponents of a base greater than $1$: $2^{-5} < 2^{-3} < 1 < 2^3$, because a more negative exponent is a smaller positive unit fraction. For a base in $(0,1)$ the reverse holds: $\bigl(\tfrac{1}{2}\bigr)^{-5} = 32 > \bigl(\tfrac{1}{2}\bigr)^{-3} = 8$.

Interval 3: $-1 < x < 0$ (negative proper fractions)

Odd powers stay negative and march toward $0$; even powers are positive and shrink toward $0$:

$$x < x^3 < x^5 < 0 < \cdots < x^6 < x^4 < x^2 < 1$$

Test $x = -\dfrac{1}{2}$:

• $x = -0.5$
• $x^2 = 0.25$
• $x^3 = -0.125$
• $x^4 = 0.0625$
• $x^5 = -0.03125$

Ordering: $-\tfrac{1}{2} < -\tfrac{1}{8} < 0 < \tfrac{1}{16} < \tfrac{1}{4}$. Square roots of $x$ itself are undefined in the reals on this interval, because $x < 0$.

Interval 4: $x < -1$ (large negative bases)

Magnitudes grow, and signs follow parity:

$$\cdots < x^5 < x^3 < x < 0 < 1 < |x| < x^2 < x^4$$

Test $x = -2$: $x^3 = -8$, $x = -2$, $x^2 = 4$, $x^4 = 16$. Even roots of $x$ are again undefined in the reals. Odd roots remain real and negative: $\sqrt[3]{-8} = -2$.

Interval Summary Table

When a stem gives only $x < 0$ without placing $x$ relative to $-1$, do not claim a single order for $|x|^2$ versus $|x|^3$. Split the case at $-1$, or pick two test values (for example $-\tfrac{1}{2}$ and $-2$) before you eliminate choices.

Classic Algebra Traps

Trap 1: $\sqrt{a^2 + b^2} \ne a + b$

A square root does not distribute over addition. For $a, b > 0$,

$$(a+b)^2 = a^2 + 2ab + b^2 > a^2 + b^2 \implies \sqrt{a^2 + b^2} < a + b.$$

Counterexample: $\sqrt{3^2 + 4^2} = 5$, while $3 + 4 = 7$.

Trap 2: $\sqrt{a - b} \ne \sqrt{a} - \sqrt{b}$

Counterexample: $\sqrt{100 - 64} = 6$, while $\sqrt{100} - \sqrt{64} = 2$.

Trap 3: $(a + b)^n \ne a^n + b^n$ (and $(ab)^n \ne a^n + b^n$)

The power of a product is $a^n b^n$. The power of a sum expands with binomial cross terms: $(a+b)^2 = a^2 + 2ab + b^2$. Dropping $2ab$ is the “freshman’s dream.” Likewise $(ab)^n$ is never $a^n + b^n$ for generic nonzero $a, b$.

Trap 4: $(a - b)^2$ is not $a^2 - b^2$

• $(a - b)^2 = a^2 - 2ab + b^2$ (perfect-square trinomial).
• $a^2 - b^2 = (a - b)(a + b)$ (difference of squares).

They agree only in special cases (for instance $b = 0$).

Trap 5: adding unlike radicals, and $\dfrac{1}{a+b} \ne \dfrac{1}{a} + \dfrac{1}{b}$

$\sqrt{2} + \sqrt{8} = \sqrt{2} + 2\sqrt{2} = 3\sqrt{2}$, but $\sqrt{2} + \sqrt{3}$ stays a sum. And $\dfrac{1}{2+2} = \dfrac{1}{4}$, while $\dfrac{1}{2} + \dfrac{1}{2} = 1$.

Domain Restrictions and Squaring Inequalities

$\sqrt{f(x)}$ is defined in the reals if and only if $f(x) \ge 0$. Example: $\sqrt{6 - 2x}$ requires $x \le 3$. Even fourth roots impose the same non-negativity; odd roots do not.

You may square both sides of $A > B$ only when both sides are known to be non-negative. Then $A > B \ge 0$ implies $A^2 > B^2$. If both sides are negative, the inequality reverses under squaring: $-2 > -5$ but $4 < 25$. Mixed or unknown signs make squaring invalid: $-4 < 3$ but $16 > 9$. On Quant, convert a radical inequality into $f(x) \ge 0$ first, then square only the non-negative sides that remain.

Directional Rules for Exponential Inequalities

For $b^{f(x)} > b^{g(x)}$:

• If $b > 1$, the exponential is increasing: $f(x) > g(x)$. Example: $2^{3x-1} > 2^{x+5} \implies 3x - 1 > x + 5 \implies x > 3$.
• If $0 < b < 1$, the exponential is decreasing: reverse the exponent inequality. Example: $\bigl(\tfrac{1}{3}\bigr)^{2x} > \bigl(\tfrac{1}{3}\bigr)^{6} \implies 2x < 6 \implies x < 3$.

Bases equal to $1$, $0$, or negatives are not usable exponential bases for these comparison rules. A negative base raised to a non-integer power is generally not real, so GMAT stems that compare $b^x$ for variable $x$ use $b > 0$.

Worked Trap and Inequality Problems

Worked Example 1: Ordering with $0 < y < 1$

Arrange $P = y^2$, $Q = \sqrt{y}$, and $R = 1/y$ in increasing order.

Pick $y = \dfrac{1}{4}$:

• $P = \dfrac{1}{16} = 0.0625$
• $Q = \dfrac{1}{2} = 0.50$
• $R = 4$

So $P < Q < R$, that is $y^2 < \sqrt{y} < \dfrac{1}{y}$. The same chain holds for every $y$ in $(0,1)$ because $y^2 < y < \sqrt{y} < 1$ while $1/y > 1$.

Worked Example 2: Do not distribute the square root

If $a$ and $b$ are positive, $a^2 + b^2 = 50$, and $ab = 7$, find $(a - b)^2$.

$$(a - b)^2 = a^2 - 2ab + b^2 = (a^2 + b^2) - 2ab = 50 - 14 = 36.$$

The false step $\sqrt{a^2 + b^2} = a + b$ would have sent you to $a + b = \sqrt{50}$, which is not needed and not equal to $a + b$.

Worked Example 3: Exponential inequality with a unit-fraction base

Solve $\bigl(\tfrac{1}{2}\bigr)^{1-x} > 8$. Rewrite $8 = 2^3 = \bigl(\tfrac{1}{2}\bigr)^{-3}$. Then $\bigl(\tfrac{1}{2}\bigr)^{1-x} > \bigl(\tfrac{1}{2}\bigr)^{-3}$. Because $0 < 1/2 < 1$, reverse: $1 - x < -3$, so $-x < -4$ and $x > 4$.