Algebra

Rationalising: multiply by 1, dressed up

To clear a surd from a denominator, multiply top and bottom by the conjugate: $\frac{1}{\sqrt7-\sqrt6} = \frac{\sqrt7+\sqrt6}{(\sqrt7-\sqrt6)(\sqrt7+\sqrt6)} = \sqrt7+\sqrt6$ since $(\sqrt7-\sqrt6)(\sqrt7+\sqrt6) = 7-6 = 1$. With a coefficient, divide by it once: $\frac{4}{\sqrt5-1} = \frac{4(\sqrt5+1)}{4} = \sqrt5+1$, and $\frac{3}{\sqrt5+\sqrt2} = \sqrt5-\sqrt2$. The number under the pair must differ by exactly what the numerator supplies.

Denesting $\sqrt{a+b\sqrt c}$

Try $\sqrt{a+b\sqrt c} = \sqrt p+\sqrt q$. Squaring: $p+q = a$ and $2\sqrt{pq} = b\sqrt c$. So find two numbers with sum $a$ and product $\frac{b^2c}{4}$: $\sqrt{15+4\sqrt{14}} = 2\sqrt2+\sqrt7 \quad (8+7 = 15,\; 2\sqrt{8\cdot7} = 4\sqrt{14})$ $\sqrt{11+6\sqrt2} = 3+\sqrt2 \quad (9+2 = 11,\; 2\sqrt9 = 6)$ If the cross term is negative, take the positive root of the reversed pair: $\sqrt{11-6\sqrt2} = 3-\sqrt2$.

The product-1 conjugate pair

When $x = p+\sqrt q$ satisfies $p^2-q = 1$, its reciprocal is the conjugate: $\frac1x = p-\sqrt q$. Then the reciprocal ladder runs for free: $x = 2+\sqrt3 \Rightarrow x+\frac1x = 4,\quad x^2+\frac1{x^2} = 14,\quad x^3+\frac1{x^3} = 52$ With $x = 5+2\sqrt6$: $\frac1x = 5-2\sqrt6$ (since $25-24=1$), so $x+\frac1x = 10$ and $x^2+\frac1{x^2} = 100-2 = 98$. For $p^2-q = m \ne 1$, the pair multiplies to $m$: then $x+\frac mx$, and you divide by $\sqrt m$ to use the ladder.

Reducing a polynomial in a surd

For $P(x)$ with $x = p+\sqrt q$, first reduce $P$ modulo the minimal quadratic $x^2-2px+(p^2-q) = 0$ — keep substituting $x^2 = 2px-(p^2-q)$ until everything is linear in $x$. Short cuts first, though: check whether the polynomial is symmetric (then the product-1 ladder applies) or whether $x^2$ itself simplifies.

Telescoping and comparison

Each term of $\frac{1}{\sqrt{k+1}+\sqrt k} = \sqrt{k+1}-\sqrt k$ telescopes: $\frac1{\sqrt2+1}+\frac1{\sqrt3+\sqrt2}+\frac1{\sqrt4+\sqrt3}+\frac1{\sqrt5+\sqrt4} = \sqrt5-1$ All interior terms cancel. For comparing surds, raise to a common power: $\sqrt2$ vs $\sqrt[3]3$ vs $\sqrt[4]4$ → compare $2^6 = 64$, $3^4 = 81$, $4^3 = 64$ → $\sqrt[3]3$ is the largest.

Common traps

• Rationalising with the wrong conjugate sign (the denominator must get a DIFFERENCE to become rational).
• Denesting with $p+q$ and $2pq$ mismatched — always re-square to check.
• Forgetting $p^2-q=1$ before claiming $\frac1x$ is the conjugate.
• Assuming nested radicals simplify — some (like $\sqrt{6+\sqrt{11}}$) do not denest neatly; fall back to squaring the whole equation.

Quick revision

• $\frac1{\sqrt a+\sqrt b} = \frac{\sqrt a-\sqrt b}{a-b}$.
• $\sqrt{a+b\sqrt c} = \sqrt p+\sqrt q$ with $p+q = a$, $2\sqrt{pq} = b\sqrt c$.
• $x = p+\sqrt q$, $p^2-q=1$ → $\frac1x = p-\sqrt q$; run the $x+\frac1x$ ladder.
• $\frac1{\sqrt{k+1}+\sqrt k} = \sqrt{k+1}-\sqrt k$ — telescopes.
• Compare surds by a common power: $2^6$ vs $3^4$ vs $4^3$.