A nested radical like \sqrt{5 + 2\sqrt{6}} looks intimidating the first time you see it. Two square roots, one sitting inside the other, no clear route through. Most students either give up or reach for a calculator. But sometimes — often enough to be worth learning — that two-deep radical collapses into a clean single-deep form like \sqrt{3} + \sqrt{2}. The trick is a single identity and a recognition test that takes about thirty seconds to run. This article gives you both.
The identity
The claim is this: under the right conditions,
for some pair of non-negative numbers c and d that you can find from a and b. The conditions come straight from squaring both sides. If the identity holds, then
Why: (\sqrt{c} + \sqrt{d})^2 = (\sqrt{c})^2 + 2\sqrt{c}\sqrt{d} + (\sqrt{d})^2 = c + 2\sqrt{cd} + d. Standard (p+q)^2 expansion.
Match the rational part to the rational part, and the radical part to the radical part:
- c + d = a
- 2\sqrt{cd} = \sqrt{b}, which squared gives 4cd = b
So denesting \sqrt{a + \sqrt{b}} reduces to finding two numbers whose sum is a and whose product is b/4. That is the entire method.
The recognition test
Given \sqrt{a + \sqrt{b}}, you want c and d with
Two numbers with a known sum and known product are roots of a quadratic — specifically, t^2 - at + \frac{b}{4} = 0. Solving gives
So the recognition test has three outcomes:
- a^2 - b is a perfect square (or at least non-negative and gives rational roots). Denest works cleanly.
- a^2 - b \geq 0 but gives irrational or fractional roots. Denest still works, just with less tidy c and d.
- a^2 - b < 0. No real c, d exist. Leave the radical as is.
The quick-check is: compute a^2 - b. If it is negative, stop. Otherwise, you can denest.
Worked example 1 — \sqrt{5 + 2\sqrt{6}}
The 2\sqrt{6} in front of the inner radical looks wrong for the identity, which expects a bare \sqrt{b}. Fix that first by pulling the 2 inside: 2\sqrt{6} = \sqrt{4 \cdot 6} = \sqrt{24}. So
Now a = 5 and b = 24. Find c, d with c + d = 5 and cd = 24/4 = 6. By inspection: c = 3, d = 2 (since 3 + 2 = 5 and 3 \cdot 2 = 6).
Verify by squaring the right side: (\sqrt{3} + \sqrt{2})^2 = 3 + 2\sqrt{6} + 2 = 5 + 2\sqrt{6}. Matches.
Worked example 2 — \sqrt{7 - 2\sqrt{10}}
Same move: 2\sqrt{10} = \sqrt{40}, so the expression is \sqrt{7 - \sqrt{40}}. Here a = 7, b = 40, and the minus sign inside means you want \sqrt{c} - \sqrt{d} (with c > d so the result is positive).
Find c, d with c + d = 7 and cd = 40/4 = 10. Quadratic: t^2 - 7t + 10 = 0, factoring to (t - 2)(t - 5) = 0. So t = 2 or t = 5. Take c = 5, d = 2:
Check: (\sqrt{5} - \sqrt{2})^2 = 5 - 2\sqrt{10} + 2 = 7 - 2\sqrt{10}. Matches.
Worked example 3 — \sqrt{3 + \sqrt{5}} (messier but still denests)
Here a = 3, b = 5. Find c, d with c + d = 3 and cd = 5/4. Quadratic: t^2 - 3t + 5/4 = 0. Discriminant = 9 - 5 = 4, so t = (3 \pm 2)/2 = 5/2 or 1/2.
Why: \sqrt{5/2} = \sqrt{5}/\sqrt{2} = \sqrt{10}/2 after rationalising. Similarly \sqrt{1/2} = \sqrt{2}/2. Sum them over the common denominator.
Verify: \left(\frac{\sqrt{10} + \sqrt{2}}{2}\right)^2 = \frac{10 + 2\sqrt{20} + 2}{4} = \frac{12 + 4\sqrt{5}}{4} = 3 + \sqrt{5}. Matches. So denesting still works — c and d are just fractions rather than integers. Whether you consider the final form "nicer" than the nested form is a matter of taste, but it is algebraically simpler in most contexts.
Worked example 4 — \sqrt{2 + \sqrt{3}}
a = 2, b = 3. c + d = 2, cd = 3/4. Quadratic: t^2 - 2t + 3/4 = 0. Discriminant = 4 - 3 = 1, so t = (2 \pm 1)/2 = 3/2 or 1/2.
This one shows up in trigonometry — it is exactly 2\cos(15°), which you may meet when computing \cos(15°) = \cos(45° - 30°) two different ways.
When it REALLY doesn't denest
Try \sqrt{1 + 2\sqrt{1000}}. Rewrite: 2\sqrt{1000} = \sqrt{4000}, so a = 1, b = 4000. Discriminant = a^2 - b = 1 - 4000 = -3999 < 0. No real c, d exist — the quadratic t^2 - t + 1000 = 0 has complex roots.
Conclusion: \sqrt{1 + 2\sqrt{1000}} cannot be denested in real numbers. Leave it as is.
The pattern is clear. If the inner \sqrt{b} is much larger than the outer a, you are probably out of luck. Denesting needs the outer rational part to dominate the inner radical part — precisely what a^2 \geq b captures.
The quick-check formula
Collecting the recipe into one line: to denest \sqrt{a + \sqrt{b}}, compute
- If a^2 - b < 0: no real denest. Stop.
- If a^2 - b \geq 0: set c, d = \dfrac{a \pm \sqrt{a^2 - b}}{2}. Then \sqrt{a + \sqrt{b}} = \sqrt{c} + \sqrt{d}.
For the minus version, \sqrt{a - \sqrt{b}}, the same c and d work, but you take \sqrt{c} - \sqrt{d} with c > d to keep the answer positive.
Where these show up
Denesting is not just a parlour trick. It appears in:
- Competition problems. RMO, INMO, and JEE problems routinely hide a nested radical and expect you to recognise the denest pattern as the first step. A problem like "simplify \sqrt{8 + 2\sqrt{15}} + \sqrt{8 - 2\sqrt{15}}" expects you to denest each term into \sqrt{5} + \sqrt{3} and \sqrt{5} - \sqrt{3}, then add to get 2\sqrt{5}.
- Trigonometric exact values. The closed forms of \sin 15°, \cos 15°, \cos 75°, and friends all involve \sqrt{2 \pm \sqrt{3}}, which denest into cleaner single-radical forms.
- Physics normal modes. Coupled oscillator frequencies often come out as \sqrt{a \pm \sqrt{b}}. Denesting gives the actual numeric value.
- Galois theory. The question of which nested radicals denest is one of the entry points into deeper algebraic number theory — but that is a story for another day.
The minus-inside variant
Everything above extends to \sqrt{a - \sqrt{b}} with one small change: the answer is \sqrt{c} - \sqrt{d} where c > d (so the result is a non-negative real number). The same quadratic t^2 - at + b/4 = 0 gives the same two roots; you just assign the larger one to c and the smaller to d.
For example, \sqrt{9 - 4\sqrt{5}} = \sqrt{9 - \sqrt{80}}. Here a = 9, b = 80, c + d = 9, cd = 20. Roots of t^2 - 9t + 20 = 0: t = 4 or t = 5. So c = 5, d = 4:
Check: (\sqrt{5} - 2)^2 = 5 - 4\sqrt{5} + 4 = 9 - 4\sqrt{5}. Matches.
Recognition drill
Run the test on each of these. Cover the answers with your hand and try before peeking.
- \sqrt{3 + 2\sqrt{2}}. Rewrite: \sqrt{3 + \sqrt{8}}. a = 3, b = 8. c + d = 3, cd = 2. Roots of t^2 - 3t + 2 = 0: t = 1, 2. Answer: \sqrt{2} + 1.
- \sqrt{7 + 4\sqrt{3}}. Rewrite: \sqrt{7 + \sqrt{48}}. a = 7, b = 48. c + d = 7, cd = 12. Roots of t^2 - 7t + 12 = 0: t = 3, 4. Answer: 2 + \sqrt{3}.
- \sqrt{10 + 2\sqrt{21}}. Rewrite: \sqrt{10 + \sqrt{84}}. a = 10, b = 84. c + d = 10, cd = 21. Roots of t^2 - 10t + 21 = 0: t = 3, 7. Answer: \sqrt{7} + \sqrt{3}.
- \sqrt{11 - 6\sqrt{2}}. Rewrite: \sqrt{11 - \sqrt{72}}. a = 11, b = 72. c + d = 11, cd = 18. Roots of t^2 - 11t + 18 = 0: t = 2, 9. Answer: 3 - \sqrt{2}.
With practice, you will stop writing down the quadratic and just spot c and d by eye. "Sum 7, product 12" reads straight off as "3 and 4" if you know your multiplication table — which is half the recognition.
Closing
Nested radicals look like they need some exotic technique. They don't. One identity, one tiny quadratic, and you are done in thirty seconds. The identity is \sqrt{a \pm \sqrt{b}} = \sqrt{c} \pm \sqrt{d} with c + d = a and 4cd = b. The quadratic is t^2 - at + b/4 = 0. If its roots are real and reasonable, denest. If the discriminant a^2 - b is negative, accept that this particular nested radical resists and move on. Most nested radicals you will see in Indian school and competition problems fall into the "denests cleanly" bucket — the rest are rare and usually flagged as such.