How do you evaluate #2x + 1- \sqrt { x - 10} = 2\sqrt { 3}#?

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Answer 1 · 2017-12-03T13:17:50

# x = (8sqrt 3 - 3 + isqrt( 167 - 16sqrt 3 ) )/8#
# x = (8sqrt 3 - 3 - isqrt( 167 - 16sqrt 3 ) )/8#

# x notinRR#, #x in CC#

Hence the first step is to rearnage and etc...;

#2x+1-2sqrt 3 = sqrt(x-10)#
#(2x+1-2sqrt 3 )^2 = x -10 #
#4x^2+x(3-8sqrt 3 ) + 23 - 4sqrt3 = 0#

Hence we can use the quadratic formula;

#b^2 - 4ac = (3-8sqrt 3)^2 - (4)(4)(23-4sqrt3)#
#= 201-48sqrt3 - 368 + 64sqrt 3 #
#= 16sqrt 3 -167#

#sqrt(b^2 - 4ac) = sqrt(16sqrt 3 -167) = isqrt(167 - 16sqrt 3)#

Using the formula, #x={ (-b+sqrt(b^2 - 4ac))/(2a) , (-b-sqrt(b^2 - 4ac))/(2a) } #

and the equation;

#4x^2+x(3-8sqrt 3 ) + 23 - 4sqrt3 = 0#

So hence yields;

# x = (8sqrt 3 - 3 + isqrt( 167 - 16sqrt 3 ) )/8#
# x = (8sqrt 3 - 3 - isqrt( 167 - 16sqrt 3 ) )/8#