What is the rationalising factor of the given number?

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

2 Answers
George C.
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#

George C.
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))#

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