Prove that the expression is an integer number sqrt(10+4sqrt6) - sqrt(10-4sqrt6)
We can consider the expression under the square root as being the result of square expanding.
10+4*sqrt6=4+2*2*sqrt6+6=(2+sqrt6)^2
10-4*sqrt6=4-2*2*sqrt6+6=(2-sqrt6)^2
Also we know that sqrt (a^2)=|a|
So,sqrt [(2+sqrt6)^2]=|(2+sqrt6)|
and sqrt [(2-sqrt6)^2]=|(2-sqrt6)|
But 2<sqrt6, so sqrt [(2-sqrt6)^2]=|-(2-sqrt6)|
We’ll substitute the results found in the expression:
|(2+sqrt6)| - |-(2-sqrt6)|=2+sqrt6+2-sqrt6=4
The calculated value is an integer number!
sqrt(10+4sqrt6) - sqrt(10-sqrt6).
Let sqrt (10+4sqrt6) = sqrt a+sqrtb. Squarig,
10+4sqrt6 = a+b+2sqr(ab)
Equating rationals , 10 =a+b. Or (a+b)^2 = 100.
Equating irrationals,
4sqr6 = 2ab. Squaring, 16*6 = 96= 4ab.
(a-b)^2 = (a+b)^2-4ab = 100-96 = 4.
Threfore a-b = sqrt4 = 2.
a+b =10
a-b = 2
Adding 2a = 12, a = 6 . b= a-2 = 4.
So sqrt(10+4sqrt6) = sqrt6+2.............(1)
Similarly
sqrt(10-4sqrt6) = sqrta - sqrtb . Squaring
a+b = 10 -4sqrt(ab). From this we get:
a+b = 10 and a - b = 2 Or sqrt(10-4sqrt6) = sqrt6-sqrt2.....(2)
Therefore
sqrt(10+4sqrt6)-sqrt(10-4sqrt6) = sqrt6+2)-(sqrt6-2) = 4.
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