Radical (surd) expressions — simplification, rationalization: Worked examples

You have met the parent laws. Now we hunt every corner of the topic. Before working a single problem, we lay out a map of all the cases — so that when a new problem appears, you can point to the cell it belongs to and already know the plan.

The scenario matrix

#	Case class	What triggers it	Master move	Example
A	Simplify a pure numeric square root	$N$ , $N$ has a square factor	pull out largest perfect square	Ex 1
B	Simplify a higher-index root	$3 N$ or $n N$	pull out largest perfect $n$ -th power	Ex 2
C	Simplify with variables	$stuff \cdot x^{k}$	split even/odd powers	Ex 3
D	Add / subtract surds	sum of roots	simplify each, combine like ones	Ex 4
E	Rationalize a single root denominator	$\frac{a}{n b}$	multiply by $n b^{n - 1}$	Ex 5
F	Rationalize a two-term denominator	$\frac{a}{c + d}$	multiply by the conjugate	Ex 6
G	Degenerate / edge input	perfect power, $0$ , or negative under even root	recognise, don't force the machine	Ex 7
H	Real-world word problem	geometry / physics phrasing	translate to a surd, simplify	Ex 8
I	Exam twist (nested + rationalize combined)	root-of-a-fraction with a binomial	chain the moves	Ex 9

A — Simplify a pure numeric square root

B — Simplify a higher-index root

C — Simplify with variables

D — Add / subtract surds

E — Rationalize a single root denominator

F — Rationalize a two-term denominator

G — Degenerate / edge inputs

Worked example Example 7 (cell G): Handle each edge case

(a) Simplify $144$ . (b) Simplify $0$ . (c) Simplify $3 - 8$ . (d) What is $- 9$ over the real numbers? Forecast: which of these is not a surd at all, and which one does not exist among real numbers?

Step 1 — (a) a perfect square. $144 = 1 2^{2} \Rightarrow 144 = 12$ Why this step? If the radicand is already a perfect power, the "extract" machine empties it completely — the result is rational, so it was never a surd. Always test for this first; it saves work.

Step 2 — (b) the zero radicand. $0 = 0 because 0^{2} = 0$ Why this step? Zero is the boundary case: it is a perfect square of itself, so the root is exactly $0$ — no leftover, no irrational part. The extract machine never chokes on $0$ ; it just returns $0$ for any index.

Step 3 — (c) negative under an odd root. $3 - 8 = - 2 because (- 2)^{3} = - 8$ Why this step? Odd roots keep the sign: an odd number of negatives is negative, so cube roots of negatives are fine and real.

Step 4 — (d) negative under an even root. $- 9 has no real value$ Why this step? Any real number squared is $\geq 0$ , so nothing real squares to $- 9$ . This case is outside the real-number playground (it needs imaginary numbers, a different topic).

Verify: $1 2^{2} = 144$ ✓, $0^{2} = 0$ ✓, $(- 2)^{3} = - 8$ ✓, and there is no real $x$ with $x^{2} = - 9$ ✓.

H — Real-world word problem (geometry)

Worked example Example 8 (cell H): A square field has area

$500 m^{2}$ . Give its exact side length in simplest surd form, then its diagonal. Forecast: guess whether the side is a "clean" number or a surd like $105$ .

Step 1. Side from area. A square of side $s$ has area $s^{2}$ , so $s = 500$ . Why this step? Area of a square is side $^{2}$ ; inverting that operation is exactly a square root.

Step 2. Simplify the surd. $500 = 100 \times 5 = 105 m$ Why this step? Same extraction rule as cell A — pull the largest perfect square $100$ out.

Step 3. Diagonal via the Pythagorean Theorem: $d = s^{2} + s^{2} = s 2$ . $d = 105 \cdot 2 = 1010 m$ Why this step? Look at the figure below: the diagonal (pink) cuts the square into two right triangles whose two legs are both the side $s$ . Pythagoras on that triangle gives hypotenuse $s 2$ .

The figure makes the surd concrete: the yellow double-arrow is the side $s = 105$ , and the pink arrow is the diagonal $d = 1010$ . Notice from the small right-angle mark in the corner why Pythagoras applies — the two legs meet at $9 0^{\circ}$ , so $d^{2} = s^{2} + s^{2}$ , and the surd $s 2$ is not a formula to memorise but the direct length of that pink line.

Verify (units + value): $s^{2} = (105)^{2} = 100 \times 5 = 500 m^{2}$ ✓. Numerically $s \approx 22.36 m$ , $d \approx 31.62 m$ , and $d / s = 2 \approx 1.414$ ✓.

I — Exam twist: nested move + rationalize combined

Worked example Example 9 (cell I): Simplify

$\frac{3}{5 + 2 3}$ into a denominator-free surd. Forecast: two dangers at once — a fraction under a root and a two-term surd on the bottom. Which do we tackle first?

Step 1. Pull the square root over the fraction (division law). $\frac{3}{5 + 2 3} = \frac{3}{5 + 2 3}$ Why this step? $a / b = a / b$ separates the numerator so we can attack the messy denominator alone.

Step 2. The cleanest route is to rationalize the fraction inside the root before taking the square root, using the conjugate machine — no guessing of hidden perfect squares needed. Rationalize $\frac{3}{5 + 2 3}$ with conjugate $5 - 23$ : $\frac{3}{5 + 2 3} \cdot \frac{5 - 2 3}{5 - 2 3} = \frac{3 ( 5 - 2 3 )}{25 - ( 2 3 ) ^{2}} = \frac{15 - 6 3}{25 - 12} = \frac{15 - 6 3}{13}$ Why this step? $(5 + 23) (5 - 23) = 25 - 4 \cdot 3 = 13$ , a clean integer, by Difference of Squares. The conjugate always works, so we never have to hope a term is a perfect square.

Step 3. Take the root back over the fraction. $\frac{15 - 6 3}{13} = \frac{15 - 6 3}{13}$ Why this step? Same division law as Step 1, now with a rational (integer) denominator underneath.

Step 4. Rationalize the remaining single-root denominator $13$ (cell E move). $\frac{15 - 6 3}{13} \cdot \frac{13}{13} = \frac{13 ( 15 - 6 3 )}{13} = \frac{195 - 78 3}{13}$ Why this step? Multiplying by $13 / 13$ turns the bottom into the integer $13$ — the required "denominator-free" form.

Verify (numeric): original $3/ (5 + 23) = 3/8.4641 \approx 0.35444 \approx 0.5954$ ; final $195 - 783 /13 = 195 - 135.10 /13 = 59.90 /13 \approx 7.740/13 \approx 0.5954$ ✓.

Recall Quick self-test (reveal after guessing)

Which cell is $\frac{2}{3 - 5}$ ? ::: Cell F — two-term denominator, use conjugate $3 + 5$ . Why can't $2 + 3$ be combined into one surd? ::: They are unlike radicals (different radicands); addition needs identical radicands, like $2 + 2$ . First thing to check on any $N$ ? ::: Whether $N$ is already a perfect power (cell G) — then it's not a surd at all. $(- 5)^{2} = ?$ ::: $∣ - 5∣ = 5$ , not $- 5$ — square roots are never negative.

Related: Equations with Radicals · Quadratic Formula · Rationalizing Complex Denominators