2.4 Radicals, Rational Expressions, and Scientific Notation

2.4 Radicals, Rational Expressions, and Scientific Notation

Radicals and rational expressions are where careless algebra becomes visibly wrong: a root that does not satisfy the original equation, a "cancellation" that was never legal, a coefficient outside the range that scientific notation permits. Every number used below is chosen to come out clean, so the whole section can be worked with pen and paper. The laws of exponents and logarithms belong to a neighbouring section; here they are used, not re-taught.

1. Simplifying Radicals

A radical is simplified when no perfect-square factor (or perfect-cube factor, for cube roots) remains under the sign.

Odd roots accept negative radicands ($\sqrt[3]{-125} = -5$) while even roots of negatives are not real numbers.

Radicals combine by addition only when the radicands match, exactly as $5k + 9k - 2k = 12k$:

$\sqrt{50} + 3\sqrt{18} - \sqrt{8} = 5\sqrt{2} + 9\sqrt{2} - 2\sqrt{2} = 12\sqrt{2}$

2. Rationalising Denominators

Answer keys are written without radicals in the denominator, so learn both moves.

• Monomial denominator — multiply top and bottom by that radical: $\dfrac{7}{\sqrt{21}} = \dfrac{7\sqrt{21}}{21} = \dfrac{\sqrt{21}}{3}$.
• Two-term denominator — multiply by the conjugate, the same two terms with the middle sign reversed, because $(a-b)(a+b) = a^2 - b^2$ destroys both radicals:

$\frac{6}{\sqrt{5} - \sqrt{2}} \cdot \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} + \sqrt{2}} = \frac{6(\sqrt{5} + \sqrt{2})}{5 - 2} = 2\left(\sqrt{5} + \sqrt{2}\right)$

Fractional exponents make some values easier than the radical form suggests: $27^{-2/3} = \dfrac{1}{\left(\sqrt[3]{27}\right)^2} = \dfrac{1}{9}$. A negative exponent inverts the base; it never makes the value negative.

3. Radical Equations and Extraneous Roots

Worked Example 1

Solve $\sqrt{x + 7} = x - 5$.

1. Square both sides: $x + 7 = x^2 - 10x + 25$.
2. Rearrange: $x^2 - 11x + 18 = 0$.
3. Factor: $(x - 9)(x - 2) = 0$, so $x = 9$ or $x = 2$.
4. Check both in the original equation. For $x = 9$: $\sqrt{16} = 4$ and $9 - 5 = 4$, so it holds. For $x = 2$: $\sqrt{9} = 3$ but $2 - 5 = -3$, so it fails.

The only solution is $x = 9$; $x = 2$ is extraneous. Squaring is not a reversible step — it erases sign information, so it can manufacture roots that were never solutions. Checking is part of the method, not optional neatness.

4. Rational Expressions

A rational expression is a ratio of polynomials. Simplify by factoring completely and cancelling common factors — never individual terms.

$\frac{x^2 - 9}{x^2 + x - 12} = \frac{(x-3)(x+3)}{(x+4)(x-3)} = \frac{x+3}{x+4}$

The excluded values come from the original denominator, so both $x = 3$ and $x = -4$ remain barred even though $(x-3)$ disappeared.

Operations follow ordinary fraction rules:

1. Multiply: factor, cross-cancel, then multiply.
2. Divide: multiply by the reciprocal of the second expression.
3. Add or subtract: build the least common denominator from the factored denominators.

Worked Example 2 — addition and a complex fraction

Combine $\dfrac{3}{x-2} + \dfrac{5}{x+1}$. The LCD is $(x-2)(x+1)$, so the sum is

$\frac{3(x+1) + 5(x-2)}{(x-2)(x+1)} = \frac{3x + 3 + 5x - 10}{(x-2)(x+1)} = \frac{8x - 7}{(x-2)(x+1)}$

Now simplify the complex fraction $\dfrac{\frac{1}{x} - \frac{1}{3}}{x - 3}$. The numerator becomes $\dfrac{3 - x}{3x}$, and dividing by $(x-3)$ gives

$\frac{3-x}{3x(x-3)} = \frac{-(x-3)}{3x(x-3)} = -\frac{1}{3x}$

Recognising that $3 - x = -(x - 3)$ is the whole trick, and it appears constantly.

5. Rational Equations

Clear denominators by multiplying every term by the LCD, then solve — but test each answer against the excluded values.

Solve $\dfrac{5}{x} - \dfrac{1}{2} = \dfrac{3}{2x}$. Multiplying through by $2x$ gives $10 - x = 3$, so $x = 7$. Since 7 is not excluded, it is valid; substituting confirms $\frac{5}{7} - \frac{1}{2} = \frac{10-7}{14} = \frac{3}{14}$.

Contrast $\dfrac{x}{x-4} = \dfrac{4}{x-4} + 3$. Multiplying by $(x-4)$ gives $x = 4 + 3(x-4) = 3x - 8$, so $x = 4$ — but $x = 4$ makes the original denominators zero. The equation has no solution, and "no solution" is a genuine answer option.

6. Scientific Notation

Scientific notation writes a number as $a \times 10^n$ where $1 \le \lvert a \rvert < 10$ and $n$ is an integer. Move the decimal point until exactly one non-zero digit stands to its left; the number of places moved is $\lvert n \rvert$, and $n$ is negative for numbers below 1.

• Multiplying: multiply the coefficients, add the exponents. $(3.0 \times 10^8)(4.5 \times 10^{-3}) = 13.5 \times 10^5$, which must be renormalised to $1.35 \times 10^6$ because 13.5 is not below 10.
• Dividing: divide the coefficients, subtract the exponents. $\dfrac{8.4 \times 10^{-2}}{2.1 \times 10^{5}} = 4.0 \times 10^{-7}$.
• Comparing: the larger exponent wins outright. $6.0 \times 10^{-4}$ beats $9.0 \times 10^{-5}$ despite the smaller coefficient, and the ratio $\frac{6.0}{9.0} \times 10^{1} \approx 6.7$ says it is nearly seven times as large.

Common Traps

• Splitting a radical across addition: $\sqrt{x^2 + 9}$ is not $x + 3$.
• Skipping the check after squaring, and keeping an extraneous root.
• Cancelling terms instead of factors: $\dfrac{x+3}{3}$ does not reduce to $x$.
• Reporting excluded values from the simplified expression only.
• Leaving an answer as $13.5 \times 10^5$ or $0.42 \times 10^{-3}$; neither coefficient is in $[1, 10)$.
• Assuming $10^{-6}$ makes a quantity negative — it makes it small.