Exercises — Radical (surd) expressions — simplification, rationalization

What we do: peek inside each radicand for a perfect-square factor (or perfect-cube for the cube root).

• (a) $20=4\times 5=2^2\times 5$ — a perfect square $2^2$ hides inside, so not simplest: $\sqrt{20}=2\sqrt{5}$.
• (b) $7$ is prime, nothing to pull out — simplest ✓.
• (c) $16=8\times 2=2^3\times 2$ — a perfect cube $2^3$ hides inside, so not simplest: $\sqrt[3]{16}=2\sqrt[3]{2}$.
• (d) $6=2\times 3$, no repeated prime — simplest ✓.

Answer: (b) and (d) are already simplest.

• Index $=3$ (the little number on the root — it says "cube root").
• Radicand $=5$ (what sits under the root).
• Rational factor $=4$ (the number multiplying the root; this makes it a mixed surd).

Step 1 — factor (WHY: pairs of primes are what a square root can pull out). $98=2\times 49=2\times 7^2$ Step 2 — extract the pair using $\sqrt{a^{2}b}=a\sqrt{b}$: $\sqrt{98}=\sqrt{7^2\times 2}=7\sqrt{2}.$ Check: $(7\sqrt{2}{)}^2=49\times 2=98$ ✓.

Step 1 — factor for triples (WHY: a cube root pulls out primes in groups of three). $54=2\times 27=2\times 3^3$ Step 2 — extract the triple using $\sqrt[3]{a^{3}b}=a\sqrt[3]{b}$: $\sqrt[3]{54}=\sqrt[3]{3^3\times 2}=3\sqrt[3]{2}.$ Check: $(3\sqrt[3]{2}{)}^3=27\times 2=54$ ✓.

What we do: multiply by $\frac{\sqrt{2}}{\sqrt{2}}$ — a sneaky "$1$" that moves the root upstairs. Why: $(\sqrt{2}{)}^2=2$ turns the denominator rational. $\frac{7}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{7\sqrt{2}}{2}.$

Step 1 — factor each piece into even powers (WHY: a square root pulls out even exponents whole). $75=5^2\times 3,x^3=x^2\cdot x,y^4=(y^2{)}^2$ Step 2 — apply $\sqrt{ab}=\sqrt{a}\sqrt{b}$ and $\sqrt{a^2}=a$: $\sqrt{75x^3{y}^4}=\sqrt{5^2\cdot x^2\cdot (y^2{)}^2\cdot 3x}=5x{y}^2\sqrt{3x}.$ Why $x,y\ge 0$ matters: $\sqrt{x^2}=|x|$ in general; the assumption lets us write $x$ not $|x|$.

Step 1 — simplify each to reveal a common surd (WHY: only like radicals — same index, same radicand — can be added). $\sqrt{50}=\sqrt{25\cdot 2}=5\sqrt{2},\sqrt{18}=\sqrt{9\cdot 2}=3\sqrt{2},\sqrt{8}=\sqrt{4\cdot 2}=2\sqrt{2}.$ Step 2 — factor out $\sqrt{2}$ exactly like collecting $x$-terms $ax+bx=(a+b)x$: $5\sqrt{2}+3\sqrt{2}-2\sqrt{2}=(5+3-2)\sqrt{2}=6\sqrt{2}.$

Step 1 — pick the conjugate of $\sqrt{5}-1$, which is $\sqrt{5}+1$. Why: $(a-b)(a+b)=a^2-b^2$ (see Difference of Squares) kills the root. Step 2 — multiply top and bottom: $\frac{4}{\sqrt{5}-1}\cdot \frac{\sqrt{5}+1}{\sqrt{5}+1}=\frac{4(\sqrt{5}+1)}{(\sqrt{5}{)}^2-1^2}=\frac{4(\sqrt{5}+1)}{5-1}=\frac{4(\sqrt{5}+1)}{4}=\sqrt{5}+1.$

What tool: Pythagorean Theorem $h^2=a^2+b^2$ — it answers "how long is the slanted side given the two legs?" Step 1 — square the legs (WHY: the theorem works with squares, and squaring a surd removes its root): $(3\sqrt{2}{)}^2=9\cdot 2=18,(\sqrt{2}{)}^2=2.$ Step 2 — add and take the root: $h=\sqrt{18+2}=\sqrt{20}.$ Step 3 — simplify $20=4\cdot 5$: $h=\sqrt{4\cdot 5}=2\sqrt{5}.$ Look at the red hypotenuse in the figure: its length is $2\sqrt{5}\approx 4.47$.

Step 1 — conjugate of $\sqrt{3}-\sqrt{2}$ is $\sqrt{3}+\sqrt{2}$. Step 2 — multiply: $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\cdot \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\frac{(\sqrt{3}+\sqrt{2}{)}^2}{(\sqrt{3}{)}^2-(\sqrt{2}{)}^2}.$ Step 3 — expand top with $(a+b{)}^2=a^2+2ab+b^2$: $(\sqrt{3}+\sqrt{2}{)}^2=3+2\sqrt{6}+2=5+2\sqrt{6}.$ Step 4 — denominator: $3-2=1$. So $=\frac{5+2\sqrt{6}}{1}=5+2\sqrt{6}.$

Step 1 — simplify the top into like surds: $\sqrt{12}=2\sqrt{3},\sqrt{27}=3\sqrt{3}\Rightarrow \text{top}=5\sqrt{3}.$ Step 2 — divide using $\frac{\sqrt{a}}{\sqrt{a}}=1$: $\frac{5\sqrt{3}}{\sqrt{3}}=5.$ A rational answer! The surds cancelled entirely.

Idea: group two terms, use one conjugate, then a second. Step 1 — treat $(\sqrt{2}+\sqrt{3})$ as a block; conjugate it against $+\sqrt{5}$, i.e. multiply by $\frac{(\sqrt{2}+\sqrt{3})-\sqrt{5}}{(\sqrt{2}+\sqrt{3})-\sqrt{5}}$: $\text{denominator}=(\sqrt{2}+\sqrt{3}{)}^2-(\sqrt{5}{)}^2=(2+2\sqrt{6}+3)-5=2\sqrt{6}.$ So far: $\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2\sqrt{6}}.$ Step 2 — rationalize the remaining $\sqrt{6}$ by multiplying by $\frac{\sqrt{6}}{\sqrt{6}}$: $\frac{(\sqrt{2}+\sqrt{3}-\sqrt{5})\sqrt{6}}{2\cdot 6}=\frac{\sqrt{12}+\sqrt{18}-\sqrt{30}}{12}.$ Step 3 — simplify $\sqrt{12}=2\sqrt{3},\sqrt{18}=3\sqrt{2}$: $=\frac{2\sqrt{3}+3\sqrt{2}-\sqrt{30}}{12}.$ Denominator is $12$ — rational ✓.

What tool: squaring both sides removes the root (see Equations with Radicals), but it can invent false solutions, so we must check. Step 1 — square: $x+7=(x+1{)}^2=x^2+2x+1$. Step 2 — rearrange to a quadratic: $0=x^2+2x+1-x-7=x^2+x-6=(x+3)(x-2).$ So $x=-3$ or $x=2$ (Quadratic Formula gives the same). Step 3 — check each in the ORIGINAL (WHY: squaring may add extraneous roots, and  ​ must be $\ge 0$):

• $x=2$: LHS $\sqrt{9}=3$, RHS $2+1=3$ ✓.
• $x=-3$: LHS $\sqrt{4}=2$, RHS $-3+1=-2$; $2\ne -2$ ✗ (RHS negative, root can't be).

Answer: $x=2$ only.

Step 1 — rationalize $a$ with conjugate $2+\sqrt{3}$: $a=\frac{1}{2-\sqrt{3}}\cdot \frac{2+\sqrt{3}}{2+\sqrt{3}}=\frac{2+\sqrt{3}}{4-3}=2+\sqrt{3}.$ Step 2 — find $1/a$. Since $a=2+\sqrt{3}$, rationalize its reciprocal: $\frac{1}{a}=\frac{1}{2+\sqrt{3}}\cdot \frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{1}=2-\sqrt{3}.$ Step 3 — add: the $\sqrt{3}$ terms cancel: $a+\frac{1}{a}=(2+\sqrt{3})+(2-\sqrt{3})=4.$ Answer: $4$, an integer ✓.