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

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Answer 1 · 2018-04-09T05:11:24

Please look below.

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)#