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)a=4334
Let b = sqrt(4/3) + sqrt(3/4)b=43+34

ab = 4/3 - 3/4ab=4334
ab = 7/12ab=712
a+b = 2sqrt(4/3)a+b=243
a+b = 4/sqrt3a+b=43

Now as simultaneous equations:
a(4/sqrt3 -a) = 7/12a(43a)=712
a^2 - (4a)/sqrt3 + 7/12 = 0a24a3+712=0

Quadratic equation:
a = (4/sqrt3 +- sqrt((4/sqrt3)^2 - 4 xx 7/12))/(2)a=43±(43)24×7122

a = (4/sqrt3 +-sqrt(16/3 - 7/3))/(2)a=43±163732

a = (4/sqrt(3) +- 3/sqrt3)/2a=43±332

a = (4/sqrt(3) - 3/sqrt3)/2a=43332

a = 1/(2sqrt3)a=123

Therefore sqrt(4/3) - sqrt(3/4) = 1/(2sqrt3)4334=123
and we also get sqrt(4/3) + sqrt(3/4) = 7/(2sqrt3)43+34=723