Question

Simplify: 2a^2 + 2root 6 ab + 36^2​

Answer 1

Answer:

How do you solve 11−62–√−−−−−−−−√ ?

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How do you solve (or “de-nest”) 11−62–√−−−−−−−−√?

One interesting way to do it is to find a quadratic equation with 11−62–√−−−−−−−−√ and its conjugate, 11+62–√−−−−−−−−√, as roots, and then use the quadratic formula to find those same roots. Why would this seemingly circular process be helpful? Because in practice, it’s easier to simplify a quadratic equation than it is to simplify nested surds.

If you start with the roots α and β of a quadratic equation, and you want to get back to the equation itself, you just set (x−α)(x−β)=0, so x2−(α+β)x+(αβ)=0. We’ll use this trick to write the quadratic equation that has those two roots.

So let’s start by letting α=11−62–√−−−−−−−−√ and β=11+62–√−−−−−−−−√ , and finding their product. Since they are positive real numbers, their product is a positive real number, so we’ll start by finding the product of their squares, and then take the positive square root of the result.

α2β2=(11−62–√)(11+62–√)=121−(36)(2)=49,

taking the positive square root,

αβ=11−62–√−−−−−−−−√×11+62–√−−−−−−−−√=7.

Now, let’s find the square of the sum in question.

(α+β)2=(11−62–√−−−−−−−−√+11+62–√−−−−−−−−√)2

=(11−62–√)+(11+62–√)+2×7

=11+11+14

=36

Again, since α and β are both positive real numbers, their sum is a positive real number. The positive square root of 36 is 6, so

α+β=6.

Now that we know the sum and product of α and β, we also know they are the roots of

x2−6x+7=0

The quadratic formula quickly tells you the roots are 3±2–√, and squaring them gives you 11±62–√, so

11−62–√−−−−−−−−√=3−2–√