Simplification

What are surds and indices?

Indices are powers: $2^5$ means 2 multiplied 5 times. Surds are roots that stay irrational, like $\sqrt{3}$ or $\sqrt[3]{5}$. Simplification questions ask you to bend both into easy shapes.

Laws of indices

• $a^m \times a^n = a^{m+n}$, $\frac{a^m}{a^n} = a^{m-n}$
• $(a^m)^n = a^{mn}$
• $a^0 = 1$, $a^{-n} = \frac{1}{a^n}$
• $a^{1/n} = \sqrt[n]{a}$, $a^{m/n} = (\sqrt[n]{a})^m$

Equation solving: make every base the same, then set the powers equal. Example: $2^{x+3} \times 4^{x-1} = 128$ → $2^{x+3+2x-2} = 2^7$ → $3x + 1 = 7$ → $x = 2$. Fractional-index values: $(32)^{2/5} = (2^5)^{2/5} = 2^2 = 4$; $(81)^{3/4} = 27$.

Rationalising the denominator

Multiply top and bottom by the conjugate (same terms, opposite middle sign): $\frac{1}{\sqrt{a}+\sqrt{b}} = \frac{\sqrt{a}-\sqrt{b}}{a-b}$ So $\frac{1}{\sqrt7 - \sqrt6} = \sqrt7 + \sqrt6$, and $\frac{1}{\sqrt9+\sqrt8} = \sqrt9 - \sqrt8$. A telescoping chain collapses to its ends. The plus-form chain: $\frac{1}{\sqrt9+\sqrt8} + \frac{1}{\sqrt8+\sqrt7} + \dots + \frac{1}{\sqrt2+1} = (\sqrt9-\sqrt8) + (\sqrt8-\sqrt7) + \dots + (\sqrt2-1) = \sqrt9 - 1 = 2.$ Watch the form: a chain of $\frac{1}{\sqrt{k}+\sqrt{k-1}}$ terms turns into $\sqrt{k}-\sqrt{k-1}$ and only the two ends survive.

Roots of surds: the a + 2√b split

To find $\sqrt{a + 2\sqrt{b}}$, find two numbers $x, y$ with $x + y = a$ and $xy = b$. Then the answer is $\sqrt{x} + \sqrt{y}$ (or $\sqrt x - \sqrt y$ when the middle sign is minus). Example: $\sqrt{11 + 6\sqrt2} = \sqrt{11 + 2\sqrt{18}}$; $x + y = 11$, $xy = 18$ → $x, y = 9, 2$ → answer $\sqrt9 + \sqrt2 = 3 + \sqrt2$.

Comparing surds

Put every surd to the power of the LCM of the root orders and compare the results. $\sqrt3, \sqrt[4]{7}, \sqrt[3]{5}$: raise to the 12th power → $3^6 = 729$, $7^3 = 343$, $5^4 = 625$. So $\sqrt3$ is the largest.

Surd plus its reciprocal

If $x = a + \sqrt{b}$, then $\frac{1}{x}$ is usually $a - \sqrt{b}$-type. Example: $x = 5 + 2\sqrt6$ → $\frac{1}{x} = 5 - 2\sqrt6$ (check: product = $25 - 24 = 1$). So $x + \frac{1}{x} = 10$. The condition is $a^2 - b = 1$ (or a perfect square): only then does the product land on a whole number. Always do this one-line check before writing the reciprocal.

Typical exam traps

• $(a^m)^n = a^{mn}$, not $a^{m+n}$: $(2^3)^2 = 64$, not $2^5$.
• Negative power flips the base: $2^{-3} = \frac{1}{8}$ — an option pair (8 vs $\frac{1}{8}$) usually tests exactly this.
• $\sqrt{a+b}$ is not $\sqrt a + \sqrt b$: $\sqrt{9+16} = 5$, not 7.
• In $\sqrt{a + 2\sqrt b}$, forget to write the middle term as $2\sqrt b$ and the split fails: $6\sqrt2 = 2\sqrt{18}$, so $b = 18$, not 2.

Quick revision

• Same base → equate exponents; fractional power = root first, then power.
• Conjugate rationalises; telescoping chains collapse to first minus last.
• $\sqrt{a+2\sqrt b}$: split $a$ into two factors of $b$.
• Compare surds via LCM-of-orders power.
• $x = a+2\sqrt b$-type → reciprocal is $a-2\sqrt b$.