The first time you meet \sqrt{3 + 2\sqrt{2}} on a JEE practice sheet, the instinct is to reach for a calculator. The radicand is a number, yes — about 5.83 — and its square root is about 2.414. But that decimal is not the answer the question wants. The expression is asking to be simplified, and "simplified" here means finding a cleaner form without a nested radical.

The surprise is that many nested radicals have beautifully simple closed forms. \sqrt{3 + 2\sqrt{2}} turns out to equal 1 + \sqrt{2} exactly — no rounding, no approximation. The technique that gets you there is called denesting, and once you see how it works, any reasonable nested radical on a JEE paper becomes a two-line problem.

The target form and the matching idea

The typical nested radical on an exam looks like

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

with a, b, c rational and c positive. The denesting question is: can this be written as \sqrt{x} + \sqrt{y} for some rationals x, y?

Here is the key move. If \sqrt{a + b\sqrt{c}} = \sqrt{x} + \sqrt{y}, squaring both sides should give an identity:

a + b\sqrt{c} = \left(\sqrt{x} + \sqrt{y}\right)^2 = x + y + 2\sqrt{xy}.

Why the squared form looks like this: (\sqrt{x}+\sqrt{y})^2 = (\sqrt{x})^2 + 2\sqrt{x}\sqrt{y} + (\sqrt{y})^2 = x + y + 2\sqrt{xy} by the standard (p+q)^2 expansion.

Now compare the two sides term by term. On the left we have a rational part a and a radical part b\sqrt{c}. On the right we have a rational part x + y and a radical part 2\sqrt{xy}. For the two sides to match, the rational parts must be equal and the radical parts must be equal. So:

\boxed{x + y = a} \qquad \boxed{xy = \frac{b^2 c}{4}}.

Two equations in two unknowns — a sum and a product — which you already know how to solve from the quadratic-equations chapter. That is the whole method.

Working through √(3 + 2√2)

Let us apply the recipe to the classic example.

Step 1. Identify a, b, c. Writing 3 + 2\sqrt{2} in the form a + b\sqrt{c}, we read off a = 3, b = 2, c = 2.

Step 2. Set up the sum and product.

x + y = 3, \qquad xy = \frac{2^2 \cdot 2}{4} = 2.

Step 3. Find x and y. You need two numbers that add to 3 and multiply to 2. Those are 1 and 2 — either from inspection or from the quadratic t^2 - 3t + 2 = 0, which factors as (t-1)(t-2) = 0.

Why "sum-and-product" works: whenever you know x + y = s and xy = p, the two unknowns are the roots of the quadratic t^2 - st + p = 0 (see Sum and Product of Roots if the connection is unfamiliar). So denesting problems hand you a ready-made factorable quadratic.

Step 4. Plug in.

\sqrt{3 + 2\sqrt{2}} = \sqrt{x} + \sqrt{y} = \sqrt{1} + \sqrt{2} = 1 + \sqrt{2}.

Step 5. Sanity check. Square the candidate: (1 + \sqrt{2})^2 = 1 + 2\sqrt{2} + 2 = 3 + 2\sqrt{2}. ✓ The decimal check agrees too: 1 + 1.414 = 2.414, and \sqrt{3 + 2(1.414)} = \sqrt{5.828} \approx 2.414.

Two lines of algebra, one quick verification, done. The calculator never had to make an appearance.

The general recipe

For any nested radical of the form \sqrt{a + b\sqrt{c}}:

  1. Solve the system x + y = a, xy = \dfrac{b^2 c}{4} for rational x and y.
  2. If the system has rational solutions, the denesting is \sqrt{a + b\sqrt{c}} = \sqrt{x} + \sqrt{y}.
  3. If the system has no rational solutions, the radical genuinely does not denest into this form — leave it alone.

The minus-sign variant works identically:

\sqrt{a - b\sqrt{c}} = \sqrt{x} - \sqrt{y} \quad\text{with}\quad x + y = a, \; xy = \frac{b^2 c}{4}, \; x > y.

The sign inside matches the sign outside because (\sqrt{x} - \sqrt{y})^2 = x + y - 2\sqrt{xy}.

A second example with a minus sign

Simplify \sqrt{7 - 4\sqrt{3}}.

Identify a = 7, b = 4, c = 3. Set up:

x + y = 7, \qquad xy = \frac{16 \cdot 3}{4} = 12.

Two numbers summing to 7 and multiplying to 12: these are 4 and 3. Order them with x > y since the original sign is minus: x = 4, y = 3.

\sqrt{7 - 4\sqrt{3}} = \sqrt{4} - \sqrt{3} = 2 - \sqrt{3}.

Verify: (2 - \sqrt{3})^2 = 4 - 4\sqrt{3} + 3 = 7 - 4\sqrt{3}. ✓

When the method fails

Not every nested radical denests into \sqrt{x} + \sqrt{y} with rational x, y. Try \sqrt{2 + \sqrt{3}}:

a = 2, b = 1, c = 3. So x + y = 2 and xy = 3/4. The quadratic t^2 - 2t + 3/4 = 0 has discriminant 4 - 3 = 1, giving t = (2 \pm 1)/2, i.e., t = 3/2 and t = 1/2. So

\sqrt{2 + \sqrt{3}} = \sqrt{\tfrac{3}{2}} + \sqrt{\tfrac{1}{2}} = \frac{\sqrt{3} + 1}{\sqrt{2}} = \frac{\sqrt{6} + \sqrt{2}}{2}.

It does denest — just not to integer radicands. The method still worked; the x and y happened to be fractions rather than integers.

But now try \sqrt{1 + \sqrt{2}}. Here x + y = 1 and xy = 2/4 = 1/2. The quadratic t^2 - t + 1/2 = 0 has discriminant 1 - 2 = -1 < 0 — no real solutions at all. So there is no way to write \sqrt{1 + \sqrt{2}} as a simple \sqrt{x} + \sqrt{y} with real x, y. It is genuinely irreducible in this form.

The discriminant test. The quadratic t^2 - a t + \tfrac{b^2 c}{4} = 0 has real roots exactly when

a^2 - b^2 c \geq 0, \quad\text{i.e.,}\quad a^2 \geq b^2 c.

So a nested radical \sqrt{a + b\sqrt{c}} denests cleanly (with x, y real) if and only if a^2 \geq b^2 c. In the successful cases above, 3^2 = 9 > 4 \cdot 2 = 8 and 7^2 = 49 > 16 \cdot 3 = 48, both just barely passing. When a^2 - b^2 c is a perfect square of a rational, the denesting is especially clean (you get integer or simple-fraction x, y).

The pattern that shows up on exams

Most JEE denesting problems are designed so a^2 - b^2 c is a nice perfect square, and x, y come out to small integers. Recognising the form \sqrt{a + b\sqrt{c}} should trigger the denesting attempt immediately — do not reach for a decimal approximation. If the discriminant check passes, you will be out with a clean answer in two lines.

A few common denestings worth memorising from practice:

These are the "greatest hits" of the topic, and having them at your fingertips means you often recognise the answer before doing the algebra.

Triple-nested and weirder radicals

Occasionally a problem gives you \sqrt{\sqrt{5}+\sqrt{21}} or \sqrt{a + \sqrt{b + \sqrt{c}}}. The same idea applies recursively: denest the innermost radical first, substitute, then denest the next layer. These are rarer and usually set up so that each layer denests nicely. If a layer fails the a^2 \geq b^2 c test, the whole tower is stuck.

For the JEE, the single-layer case \sqrt{a + b\sqrt{c}} is the dominant pattern. Master that, and you have covered 95% of the denesting questions that appear.

Related: Roots and Radicals · Sum and Product of Roots · Rationalise the Denominator: Watch the Conjugate Flush the Irrational Upstairs · Why √a · √b = √(ab) But √a + √b ≠ √(a+b)