Math: (Shortcut for) Simplifying (Denesting) Nested Square Roots

The basic form looks like this:

$$√{a ± 2√b}$$

a is $(x+y)$ and b is $(x·y)$

$$√{(x + y) ± 2√(xy)} = √x ± √y$$

Let's go to the methods and examples.

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PLUS

$$√{a + 2√b}$$

$$√{(x + y) + 2√(xy)} = √x + √y$$

With $x > 0$ and $y > 0$

Example

$√{7 + 2√12}$

We have $x + y = 7$ and $xy = 12$

Factor pairs of $12$:

• $1 × 12$
• $2 × 6$
• $3 × 4$

We pick the last factor pair: $3 × 4$, because $3 + 4 = 7$

We have $x = 3$ and $y = 4$

Therefore, $√{7 + 2√12}$ can be simplified to $√3 + √4$ ➡️ $√3 + 2$

In conclusion:

$√{7 + 2√12} = √3 + 2$ ✅

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MINUS

$$√{a - 2√b}$$

$$√{(x + y) - 2√(xy)} = √x - √y$$

With $x > 0$, $y > 0$, and $x > y$

Example

$√{15 - 2√56}$

We have $x + y = 15$ and $xy = 56$

Factor pairs of $56$:

• $1 × 56$
• $2 × 28$
• $4 × 14$
• $7 × 8$

We pick the last factor pair: $7 × 8$, because $7 + 8 = 15$

But since $x > y$, then we put the $x = 8$ and $y = 7$

Therefore, $√{15 - 2√56}$ can be simplified to $√8 - √7$ ➡️ $2√2 - √7$

In conclusion:

$√{15 - 2√56} = 2√2 - √7$ ✅

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Tricks: Radical Denesting Using Binomial Square Identity 🐴

That probably looks like mathematical jargons put together. But what it's trying to say is:

We're simplifying square root expressions using the identity:

$${(√x ± √y)}² = x + y ± 2√{(xy)}$$

From the basic identity:

$${(p ± q)}² = p² + q² ± 2√{(pq)}$$

You see, the form at the beginning of this article stems from this identity and we take the square root on both sides:

$$√{{(√x ± √y)}²} = √{x + y ± 2√{(xy)}}$$

Final form:

$$√x ± √y = √{(x + y) ± 2√{(xy)}}$$

⬆️ Similar to the beginning of this article.

Approved by Euler's ghost

Hm. Let's go then.

COMPRESSION

$√{12 + 4√8}$

We need to "compress" the $4$ to be $2$!

As such:

$√{12 + 4√8} = √{12 + 2·2√8} = √{12 + 2√{4·8}} = √{12 + 2√32}$

So now we have, $√{12 + 4√8} = √{12 + 2√32}$

Using the method from the examples above, we will get:

$√{12 + 2√32} = √8 + √4 = 2√2 + 2 = 2(√2 + 1)$ ✅

EXPANSION

$√{9 - √72}$

We need to summon the "2 coefficient" by extracting it from the $√72$! In other words, expand the 1 coefficient to 2.

Like so:

$√{9 - √72} = √{9 - √{4·18}} = √{9 - 2√18$

And then we have, $√{9 - √72} = √{9 - 2√18}$

Using the method from the examples above, we will get:

$√{9 - √72} = √{9 - 2√18} = √6 - √3$ ✅

🥳