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