What is the rationalising factor of the given number?

2+sqrt(7+2sqrt(10)2+7+210

2 Answers
Aug 13, 2018

6+sqrt(5)+7sqrt(2)-4sqrt(10)6+5+72410

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))2+7+210

First we should check whether the expression sqrt(7+2sqrt(10))7+210 can be simplified.

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

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

then if p^2-q^2rp2q2r is a perfect square s^2s2 then:

sqrt(p+qsqrt(r)) = sqrt(2p+2s)/2+sqrt(2p-2s)/2p+qr=2p+2s2+2p2s2

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

s = sqrt(p^2-q^2r) = sqrt(7^2-2^2(10)) = sqrt(49-40) = sqrt(9) = 3s=p2q2r=7222(10)=4940=9=3

So:

sqrt(7+2sqrt(10)) = sqrt(2(7)+2(3))/2+sqrt(2(7)-2(3))/27+210=2(7)+2(3)2+2(7)2(3)2

color(white)(sqrt(7+2sqrt(10))) = sqrt(20)/2+sqrt(8)/27+210=202+82

color(white)(sqrt(7+2sqrt(10))) = sqrt(5)+sqrt(2)7+210=5+2

So:

2+sqrt(7+2sqrt(10)) = 2+sqrt(5)+sqrt(2)2+7+210=2+5+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)2±5±2 apart from the one we already have.

(2+sqrt(5)-sqrt(2))(2-sqrt(5)-sqrt(2))(2-sqrt(5)+sqrt(2))(2+52)(252)(25+2)

=(2+sqrt(5)-sqrt(2))((2-sqrt(5))^2-(sqrt(2))^2)=(2+52)((25)2(2)2)

=(2+sqrt(5)-sqrt(2))((4-4sqrt(5)+5)-2)=(2+52)((445+5)2)

=(2+sqrt(5)-sqrt(2))(7-4sqrt(5))=(2+52)(745)

=7(2+sqrt(5)-sqrt(2))-4sqrt(5)(2+sqrt(5)-sqrt(2))=7(2+52)45(2+52)

=(14+7sqrt(5)-7sqrt(2))-(8sqrt(5)+20-4sqrt(10))=(14+7572)(85+20410)

=-6-sqrt(5)-7sqrt(2)+4sqrt(10)=6572+410

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)6+5+72410

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

(2+sqrt(5)+sqrt(2))(6+sqrt(5)+7sqrt(2)-4sqrt(10))(2+5+2)(6+5+72410)

=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))=2(6+5+72410)+5(6+5+72410)+2(6+5+72410)

=(12+2sqrt(5)+14sqrt(2)-8sqrt(10))+(6sqrt(5)+5+7sqrt(10)-20sqrt(2))+(6sqrt(2)+sqrt(10)+14-8sqrt(5))=(12+25+142810)+(65+5+710202)+(62+10+1485)

=31=31

Aug 13, 2018

Here's another way to simplify...

Explanation:

One way of simplifying sqrt(7+2sqrt(10))7+210 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))7+210, " "sqrt(7-2sqrt(10)) 7210, " " and " "-sqrt(7-2sqrt(10)) 7210

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

(x^2-7)^2-40(x27)240

= x^4-14x^2+9=x414x2+9

= x^4-6x^2+9-8x^2=x46x2+98x2

= (x^2-3)^2-(2sqrt(2)x)^2=(x23)2(22x)2

= (x^2-2sqrt(2)x-3)(x^2+2sqrt(2)x-3)=(x222x3)(x2+22x3)

= (x^2-2sqrt(2)x+2-5)(x^2+2sqrt(2)x+2-5)=(x222x+25)(x2+22x+25)

= ((x-sqrt(2))^2-(sqrt(5))^2)((x+sqrt(2))^2-(sqrt(5))^2)=((x2)2(5)2)((x+2)2(5)2)

= (x-sqrt(2)-sqrt(5))(x-sqrt(2)+sqrt(5))(x+sqrt(2)-sqrt(5))(x+sqrt(2)+sqrt(5))=(x25)(x2+5)(x+25)(x+2+5)

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

The greatest of these is x=sqrt(2)+sqrt(5)x=2+5, which must be the greatest of +-sqrt(7+-2sqrt(10))±7±210, i.e. sqrt(7+2sqrt(10))7+210

So:

sqrt(7+2sqrt(10)) = sqrt(2)+sqrt(5)7+210=2+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))(2+52)(252)(25+2)