How do you evaluate square root of 4/3 minus the square root of 3/4?

1 Answer
Apr 9, 2018

Please look below.

Explanation:

Let a = sqrt(4/3) - sqrt(3/4)
Let b = sqrt(4/3) + sqrt(3/4)

ab = 4/3 - 3/4
ab = 7/12
a+b = 2sqrt(4/3)
a+b = 4/sqrt3

Now as simultaneous equations:
a(4/sqrt3 -a) = 7/12
a^2 - (4a)/sqrt3 + 7/12 = 0

Quadratic equation:
a = (4/sqrt3 +- sqrt((4/sqrt3)^2 - 4 xx 7/12))/(2)

a = (4/sqrt3 +-sqrt(16/3 - 7/3))/(2)

a = (4/sqrt(3) +- 3/sqrt3)/2

a = (4/sqrt(3) - 3/sqrt3)/2

a = 1/(2sqrt3)

Therefore sqrt(4/3) - sqrt(3/4) = 1/(2sqrt3)
and we also get sqrt(4/3) + sqrt(3/4) = 7/(2sqrt3)