What is the rationalising factor of the given number?
2+sqrt(7+2sqrt(10)2+√7+2√10
2 Answers
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+2√10
First we should check whether the expression
Note that in https://socratic.org/s/aTtiPKas I found that given:
sqrt(p+qsqrt(r))" "√p+q√r withp, q, r > 0p,q,r>0
then if
sqrt(p+qsqrt(r)) = sqrt(2p+2s)/2+sqrt(2p-2s)/2√p+q√r=√2p+2s2+√2p−2s2
In our example, putting
s = sqrt(p^2-q^2r) = sqrt(7^2-2^2(10)) = sqrt(49-40) = sqrt(9) = 3s=√p2−q2r=√72−22(10)=√49−40=√9=3
So:
sqrt(7+2sqrt(10)) = sqrt(2(7)+2(3))/2+sqrt(2(7)-2(3))/2√7+2√10=√2(7)+2(3)2+√2(7)−2(3)2
color(white)(sqrt(7+2sqrt(10))) = sqrt(20)/2+sqrt(8)/2√7+2√10=√202+√82
color(white)(sqrt(7+2sqrt(10))) = sqrt(5)+sqrt(2)√7+2√10=√5+√2
So:
2+sqrt(7+2sqrt(10)) = 2+sqrt(5)+sqrt(2)2+√7+2√10=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-sqrt(5)-sqrt(2))(2-sqrt(5)+sqrt(2))(2+√5−√2)(2−√5−√2)(2−√5+√2)
=(2+sqrt(5)-sqrt(2))((2-sqrt(5))^2-(sqrt(2))^2)=(2+√5−√2)((2−√5)2−(√2)2)
=(2+sqrt(5)-sqrt(2))((4-4sqrt(5)+5)-2)=(2+√5−√2)((4−4√5+5)−2)
=(2+sqrt(5)-sqrt(2))(7-4sqrt(5))=(2+√5−√2)(7−4√5)
=7(2+sqrt(5)-sqrt(2))-4sqrt(5)(2+sqrt(5)-sqrt(2))=7(2+√5−√2)−4√5(2+√5−√2)
=(14+7sqrt(5)-7sqrt(2))-(8sqrt(5)+20-4sqrt(10))=(14+7√5−7√2)−(8√5+20−4√10)
=-6-sqrt(5)-7sqrt(2)+4sqrt(10)=−6−√5−7√2+4√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)6+√5+7√2−4√10
As a check, let's multiply this by
(2+sqrt(5)+sqrt(2))(6+sqrt(5)+7sqrt(2)-4sqrt(10))(2+√5+√2)(6+√5+7√2−4√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))=2(6+√5+7√2−4√10)+√5(6+√5+7√2−4√10)+√2(6+√5+7√2−4√10)
=(12+2sqrt(5)+14sqrt(2)-8sqrt(10))+(6sqrt(5)+5+7sqrt(10)-20sqrt(2))+(6sqrt(2)+sqrt(10)+14-8sqrt(5))=(12+2√5+14√2−8√10)+(6√5+5+7√10−20√2)+(6√2+√10+14−8√5)
=31=31
Here's another way to simplify...
Explanation:
One way of simplifying
The other zeros will be the variants:
-sqrt(7+2sqrt(10))−√7+2√10 ," "sqrt(7-2sqrt(10)) √7−2√10 ," " and" "-sqrt(7-2sqrt(10)) −√7−2√10
The quartic of which these are zeros can be expressed as:
(x^2-7)^2-40(x2−7)2−40
= x^4-14x^2+9=x4−14x2+9
= x^4-6x^2+9-8x^2=x4−6x2+9−8x2
= (x^2-3)^2-(2sqrt(2)x)^2=(x2−3)2−(2√2x)2
= (x^2-2sqrt(2)x-3)(x^2+2sqrt(2)x-3)=(x2−2√2x−3)(x2+2√2x−3)
= (x^2-2sqrt(2)x+2-5)(x^2+2sqrt(2)x+2-5)=(x2−2√2x+2−5)(x2+2√2x+2−5)
= ((x-sqrt(2))^2-(sqrt(5))^2)((x+sqrt(2))^2-(sqrt(5))^2)=((x−√2)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))=(x−√2−√5)(x−√2+√5)(x+√2−√5)(x+√2+√5)
Hence zeros:
The greatest of these is
So:
sqrt(7+2sqrt(10)) = sqrt(2)+sqrt(5)√7+2√10=√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+√5−√2)(2−√5−√2)(2−√5+√2)