What is the rationalising factor of the given number?

2+sqrt(7+2sqrt(10)

2 Answers
Aug 13, 2018

6+sqrt(5)+7sqrt(2)-4sqrt(10)

Explanation:

I will assume that you are looking for a radical conjugate, that is an expression which when multiplied by the given expression gives a rational product.

Given:

2+sqrt(7+2sqrt(10))

First we should check whether the expression sqrt(7+2sqrt(10)) can be simplified.

Note that in https://socratic.org/s/aTtiPKas I found that given:

sqrt(p+qsqrt(r))" " with p, q, r > 0

then if p^2-q^2r is a perfect square s^2 then:

sqrt(p+qsqrt(r)) = sqrt(2p+2s)/2+sqrt(2p-2s)/2

In our example, putting p = 7, q = 2, r = 10, then:

s = sqrt(p^2-q^2r) = sqrt(7^2-2^2(10)) = sqrt(49-40) = sqrt(9) = 3

So:

sqrt(7+2sqrt(10)) = sqrt(2(7)+2(3))/2+sqrt(2(7)-2(3))/2

color(white)(sqrt(7+2sqrt(10))) = sqrt(20)/2+sqrt(8)/2

color(white)(sqrt(7+2sqrt(10))) = sqrt(5)+sqrt(2)

So:

2+sqrt(7+2sqrt(10)) = 2+sqrt(5)+sqrt(2)

Since this involves two square roots, a suitable radical conjugate is formed by multiplying the variants of the expression 2+-sqrt(5)+-sqrt(2) apart from the one we already have.

(2+sqrt(5)-sqrt(2))(2-sqrt(5)-sqrt(2))(2-sqrt(5)+sqrt(2))

=(2+sqrt(5)-sqrt(2))((2-sqrt(5))^2-(sqrt(2))^2)

=(2+sqrt(5)-sqrt(2))((4-4sqrt(5)+5)-2)

=(2+sqrt(5)-sqrt(2))(7-4sqrt(5))

=7(2+sqrt(5)-sqrt(2))-4sqrt(5)(2+sqrt(5)-sqrt(2))

=(14+7sqrt(5)-7sqrt(2))-(8sqrt(5)+20-4sqrt(10))

=-6-sqrt(5)-7sqrt(2)+4sqrt(10)

Note that this is negative, so, let's negate it to get the slightly more attractive radical conjugate:

6+sqrt(5)+7sqrt(2)-4sqrt(10)

As a check, let's multiply this by 2+sqrt(5)+sqrt(2) and see what we get:

(2+sqrt(5)+sqrt(2))(6+sqrt(5)+7sqrt(2)-4sqrt(10))

=2(6+sqrt(5)+7sqrt(2)-4sqrt(10))+sqrt(5)(6+sqrt(5)+7sqrt(2)-4sqrt(10))+sqrt(2)(6+sqrt(5)+7sqrt(2)-4sqrt(10))

=(12+2sqrt(5)+14sqrt(2)-8sqrt(10))+(6sqrt(5)+5+7sqrt(10)-20sqrt(2))+(6sqrt(2)+sqrt(10)+14-8sqrt(5))

=31

Aug 13, 2018

Here's another way to simplify...

Explanation:

One way of simplifying sqrt(7+2sqrt(10)) involves considering the quartic polynomial with integer coefficients of which it is one of the zeros.

The other zeros will be the variants:

-sqrt(7+2sqrt(10)), " "sqrt(7-2sqrt(10)), " " and " "-sqrt(7-2sqrt(10))

The quartic of which these are zeros can be expressed as:

(x^2-7)^2-40

= x^4-14x^2+9

= x^4-6x^2+9-8x^2

= (x^2-3)^2-(2sqrt(2)x)^2

= (x^2-2sqrt(2)x-3)(x^2+2sqrt(2)x-3)

= (x^2-2sqrt(2)x+2-5)(x^2+2sqrt(2)x+2-5)

= ((x-sqrt(2))^2-(sqrt(5))^2)((x+sqrt(2))^2-(sqrt(5))^2)

= (x-sqrt(2)-sqrt(5))(x-sqrt(2)+sqrt(5))(x+sqrt(2)-sqrt(5))(x+sqrt(2)+sqrt(5))

Hence zeros: x = +-sqrt(2)+-sqrt(5)

The greatest of these is x=sqrt(2)+sqrt(5), which must be the greatest of +-sqrt(7+-2sqrt(10)), i.e. sqrt(7+2sqrt(10))

So:

sqrt(7+2sqrt(10)) = sqrt(2)+sqrt(5)

From this point we can proceed as my other answer to multiply:

(2+sqrt(5)-sqrt(2))(2-sqrt(5)-sqrt(2))(2-sqrt(5)+sqrt(2))