3. Surds

🧠 What are Surds About?

Surds let us write exact irrational values like $2\sqrt{3}$ and $\sqrt{5}$ instead of rounding them to decimals. In A-Math, you must manipulate surds confidently and leave answers in exact form when required.

This chapter focuses on:

• understanding what a surd is and when to use exact form (surds),
• simplifying and combining expressions with surds,
• rationalising denominators (single-term and binomial), and
• solving equations that involve square roots safely (avoid extraneous roots).

📌 E-Math often accepts decimals. A-Math expects exact surd form unless the question asks for decimals.

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🔹 3.1 Surds

🧠 Key Concepts

• A surd is an irrational root such as $\sqrt{2}$, $\sqrt[3]{3}$, $\sqrt[7]{5}$, or expressions involving them like $3\sqrt{7}$.
• Simplest surd form: $k\sqrt{n}$ where $n$ has no square factors $>1$ (i.e., $n$ is square-free) and $k$ is rational.

🔹 3.2 Simplifying Expressions Involving Surds

🧠 Key Concepts

🤓 3 Laws of Surds:

1. Times: $\sqrt{ab}=\sqrt{a}\times \sqrt{b}$
2. Divide: $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$ $(b>0)$
3. Squaring square roots: $\sqrt{a^2}=|a|$ (use absolute value when variables may be negative)

🤓 Conjugate surds

• Pair of binomial expressions $a+\sqrt{b}$ and $a-\sqrt{b}$, where $a$ and $b$ are integers
• Special identity: $(a+\sqrt{b})(a-\sqrt{b})=a^2-b^2$
• Mainly used for “rationalising the denominator”
• Conjugates: The conjugate of $a+\sqrt{b}$ is $a-\sqrt{b}$, and vice versa.

Rationalising denominators

• Goal: Making fractions with irrational denominators (containing surds) have rational denominators (e.g. $\frac{5}{6-\sqrt{7}}=\frac{5}{6-\sqrt{7}}\times \frac{6+\sqrt{7}}{6+\sqrt{7}}=\frac{5(6+\sqrt{7})}{6^2-7}=\frac{5(6+\sqrt{7})}{29}$)

🏷️ Definitions & Terminology

• Surd
  • Irrational root of numbers which cannot simplify into whole numbers / exact fractions
• Square-free
  • Integer with no square factor $>1$ (e.g., $12$ is not square-free because $12=4\times 3$).
• Conjugate surds
  • For a single surd $\sqrt{x}$, its conjugates are $\sqrt{x}$ and $-\sqrt{x}$
  • For a binomial surd $a+\sqrt{b}$, its conjugate is $a-\sqrt{b}$
• Exact form
  • Leave answers in surd form, not rounded decimals, unless specified.

📐 Other Concepts Tested

🤓 Adding/Subtracting Surds

• Example: $\sqrt{32}+\sqrt{50}$
• Idea: Factor into squares and non-squares
• Explanation
  • $32$ and $50$ are both factors of 2
  • $\sqrt{32}=\sqrt{16\times 2}=\sqrt{4^2\times 2}=4\sqrt{2}$
  • $\sqrt{50}=\sqrt{25\times 2}=\sqrt{5^2\times 2}=5\sqrt{2}$
  • Next, we add them together: $\sqrt{32}+\sqrt{50}=4\sqrt{2}+5\sqrt{2}=9\sqrt{2}$

Multiplying Binomial Surds

• Example: $(3+5\sqrt{2})(4-\sqrt{2})$
• Idea: Use the Umbrella method
• Solution \begin{align}(3+5\sqrt{2})(4-\sqrt{2}) &= (12 - 3\sqrt{2} + 20\sqrt{2} - 10) \text{ (umbrella method)}\\ &= 2 + 17\sqrt{2}\end{align}

🤓 Rationalising denominators

• Single surd: $\frac{a}{\sqrt{b}}=\frac{a\sqrt{b}}{b}$.
• Binomial: $\frac{1}{p+\sqrt{q}}$ ⇒ multiply top and bottom by conjugate $(p-\sqrt{q})$.

Tips

• Treat surds like algebra: $2\sqrt{5}+3\sqrt{5}=5\sqrt{5}$ (similar to $2x+3x=5x$)
• Do prime factorisation to extract perfect squares
• Dividing by surd $⟹$ Rationalising denominators

Worked Examples

Question: Simplify $\sqrt{50}$.

• Step 1: Factor $50=25\times 2$.
• Step 2: $\sqrt{50}=\sqrt{25\times 2}=\sqrt{25}\sqrt{2}=5\sqrt{2}$.
• Step 3: Final answer: $5\sqrt{2}$ (square-free inside the root).

Question: Rationalise and simplify $\frac{4}{\sqrt{3}}$.

• Step 1: Multiply by $\frac{\sqrt{3}}{\sqrt{3}}$: $\frac{4}{\sqrt{3}}=\frac{4\sqrt{3}}{3}$
• Step 2: Final answer: $\frac{4\sqrt{3}}{3}$

Question: Rationalise and simplify $\frac{3}{2+\sqrt{5}}$.

• Step 1: Multiply by conjugate $\frac{2-\sqrt{5}}{2-\sqrt{5}}$
• Step 2: Numerator $3(2-\sqrt{5})=6-3\sqrt{5}$
• Step 3: Denominator $(2+\sqrt{5})(2-\sqrt{5})=4-5=-1$
• Step 4: Result $=\frac{6-3\sqrt{5}}{-1}=3\sqrt{5}-6$

Question: Express $\sqrt{12x^3}$ in simplest surd form, given $x>0$.

• Step 1: Factor $12x^3=4\times 3\times x^2\times x$
• Step 2: $\sqrt{12x^3}=\sqrt{4}\sqrt{3}\sqrt{x^2}\sqrt{x}=2\cdot \sqrt{3}\cdot x\cdot \sqrt{x}=2x\sqrt{3x}$
• Step 3: Final answer: $2x\sqrt{3x}$

Question: Simplify $(\sqrt{8}-\sqrt{2}{)}^2$.

• Step 1: Expand: $(\sqrt{8}{)}^2-2\sqrt{8}\sqrt{2}+(\sqrt{2}{)}^2=8-2\sqrt{16}+2$.
• Step 2: $\sqrt{16}=4$, so expression $=8-8+2=2$.
• Step 3: Final answer: $2$.

Question: Expand and simplify $(3+\sqrt{5})(2-\sqrt{5})$.

• Step 1: Distribute: $3\cdot 2+3(-\sqrt{5})+\sqrt{5}\cdot 2+\sqrt{5}(-\sqrt{5})$.
• Step 2: $=6-3\sqrt{5}+2\sqrt{5}-(\sqrt{5}{)}^2=6-\sqrt{5}-5$.
• Step 3: Final answer: $1-\sqrt{5}$.

Question: Simplify $\frac{\sqrt{18}+\sqrt{8}}{\sqrt{2}}$.

• Step 1: Simplify surds: $\sqrt{18}=3\sqrt{2}$, $\sqrt{8}=2\sqrt{2}$.
• Step 2: Numerator $=(3\sqrt{2}+2\sqrt{2})=5\sqrt{2}$.
• Step 3: Divide by $\sqrt{2}$: $\frac{5\sqrt{2}}{\sqrt{2}}=5$.

Question: Simplify $\frac{\sqrt{27x^{2}y}}{\sqrt{3y}}$, given $x,y>0$.

• Step 1: Combine under one root: $\sqrt{\frac{27x^{2}y}{3y}}=\sqrt{9x^2}=3x$.
• Step 2: Final answer: $3x$.

🎯 Exam Thinking

• Is the final surd square-free inside $\sqrt{\cdot }$?
  • Square-free: Contains a square factor
• Did I remove all surds from the denominator?

❌ Common Mistakes ⚠️

• Adding unlike surds without simplification
  • e.g., $\sqrt{12}+\sqrt{27}=\sqrt{39}$ — wrong
• Leaving surd in denominator
  • e.g. $\frac{5}{7-13\sqrt{5}}$ — Not simplified
• Dropping the middle term when expanding $(a+b{)}^2$
  • e.g. $(x+y{)}^2=x^2+y^2$ — $2xy$ is missing!

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🔹 3.3 Solving Equations Involving Surds

🧠 Key Concepts

🤓 Equality of Surds

• If $a+b\sqrt{n}=p+q\sqrt{n}$, then $a=p$ and $b=q$ (given $a,b,n,p,q$ are rational)
• Example: $a-5\sqrt{3}=7+b\sqrt{3}$
  • $a=7$ and $b=5$ by equality of surds
• This is used a lot in word problems involving surds!

🤓 Squaring binomial surds

• If you square binomial surds, you end up with another binomial surd!
  • e.g. $(5+6\sqrt{2}{)}^2=25+12\sqrt{2}+72=97+12\sqrt{2}$
• Scary at first, but simple once you get it!

📐 Understanding the Parameters

• If $\sqrt{f(x)}=g(x)$, then require $g(x)\ge 0$ and $f(x)\ge 0$ before solving.
• When two surds are present, consider isolating one: $\sqrt{f(x)}=h(x)-\sqrt{g(x)}$; then square, isolate again, square again.

✏️ Worked Examples

Question: Solve $\sqrt{x+3}=x-1$.

• Step 1: Square both sides \begin{align}x + 3 &= (x - 1)^2\\ &= x^2 - 2x + 1\end{align}
• Step 2: Isolate x $x^2-3x-2=0$
• Step 3: Solve for x $x=\frac{3\pm \sqrt{9+8}}{2}=\frac{3\pm \sqrt{17}}{2}$
• Step 4: Check answers
  • For $x=\frac{3+\sqrt{17}}{2}\approx 3.56$, LHS $\approx \sqrt{6.56}\approx 2.56$, RHS $\approx 2.56$ ✓.
  • For $x=\frac{3-\sqrt{17}}{2}<1$, violates domain ✗.
• Step 4: Final answer: $x=\frac{3+\sqrt{17}}{2}$.

Question: Solve $\sqrt{2x-1}+\sqrt{x-2}=3$.

• Step 1: Domain: $2x-1\ge 0\Rightarrow x\ge \frac{1}{2}$; $x-2\ge 0\Rightarrow x\ge 2$ ⇒ overall $x\ge 2$.
• Step 2: Isolate one surd: $\sqrt{2x-1}=3-\sqrt{x-2}$.
• Step 3: Square: $2x-1=9-6\sqrt{x-2}+(x-2)$.
• Step 4: Simplify: $2x-1=x+7-6\sqrt{x-2}\Rightarrow x-8=-6\sqrt{x-2}$.
• Step 5: Square again: $(x-8{)}^2=36(x-2)\Rightarrow x^2-16x+64=36x-72$.
• Step 6: Rearrange: $x^2-52x+136=0$.
• Step 7: Solve: $D=2704-544=2160=144\times 15\Rightarrow \sqrt{D}=12\sqrt{15}$.
• Step 8: $x=\frac{52\pm 12\sqrt{15}}{2}=26\pm 6\sqrt{15}$.
• Step 9: Check domain $x\ge 2$; $26-6\sqrt{15}\approx 1.78<2$ ✗; $26+6\sqrt{15}\gg 2$ ✓.
• Step 10: Final answer: $x=26+6\sqrt{15}$.

Question: Solve $3x=2+\sqrt{5x+1}$.

• Step 1: Domain: $5x+1\ge 0\Rightarrow x\ge -\frac{1}{5}$.
• Step 2: Isolate surd: $\sqrt{5x+1}=3x-2$ (so require $3x-2\ge 0\Rightarrow x\ge \frac{2}{3}$).
• Step 3: Square: $5x+1=(3x-2{)}^2=9x^2-12x+4$.
• Step 4: Rearrange: $0=9x^2-17x+3$.
• Step 5: Solve: $x=\frac{17\pm \sqrt{289-108}}{18}=\frac{17\pm \sqrt{181}}{18}$.
• Step 6: Domain check: $\frac{17-\sqrt{181}}{18}\approx 0.195<\frac{2}{3}$ ✗; $\frac{17+\sqrt{181}}{18}\approx 1.69\ge \frac{2}{3}$ ✓.
• Step 7: Final answer: $x=\frac{17+\sqrt{181}}{18}$.

🎯 Exam Thinking

• Have I set domain conditions before squaring?
• After solving, did I substitute back to reject extraneous roots?

❌ Common Mistakes ⚠️

• Squaring too early without isolating a surd first — leads to messy algebra and errors.
• Forgetting that $\sqrt{\cdot }$ is non-negative; solutions that make RHS negative are invalid.
• Not checking solutions back in the original equation.

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✅ Quick Strategy 🎯

1. Simplify each surd to square-free form before combining.
2. Combine like surds; use conjugates to eliminate roots quickly.
3. Always rationalise denominators in final answers unless told otherwise.
4. For equations, isolate a surd before squaring and check for extraneous roots.
5. State and use domain conditions early to avoid invalid answers.