# Prove that the expression is an integer number sqrt(10+4sqrt6) - sqrt(10-4sqrt6)

hala718 | Certified Educator

sqrt(10 + 4sqrt6) -sqrt(10-4sqrt6)

Let us rewrite:

10 + 4sqrt6 = (2+ sqrt6)^2

10 -4sqrt6 = (2 -sqrt6)^2

sqrt(10+4sqrt6)-sqrt(10-4sqrt6)= sqrt(2+sqrt6)^2 - sqrt(2-sqrt6)^2

= 2+ sqrt6 - (-2+sqrt6)

= 2+ sqrt6 + 2 -sqrt6

= 4

=

giorgiana1976 | Student

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!

neela | Student

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.