How do you solve #2sqrtx +1= 6/sqrtx#?

1 Answer
Jul 18, 2015

Answer:

Reformulate as a quadratic with possible spurious solutions and solve to find #x = 9/4# (and spurious solution #x=4#)

Explanation:

Multiply both sides by #sqrt(x)# to get:

#2x+sqrt(x) = 6#

Subtract #2x# from both sides to get:

#sqrt(x) = 6-2x#

Square both sides to get:

#x = (6-2x)^2 = 36 - 24x + 4x^2#

Note that squaring may introduce spurious solutions, so need to check later.

Subtract #x# from both sides to get:

#4x^2-25x+36 = 0#

Use the quadratic formula to find:

#x = (25+-sqrt(25^2-(4xx4xx36)))/(2*4)#

#= (25+-sqrt(625 - 576))/8#

#= (25+-sqrt(49))/8#

#= (25+-7)/8#

That is #x=4# or #x=9/4#

If #x=4# then:

#2sqrt(x) + 1 = 2sqrt(4) + 1 = 4 + 1 = 5#

But #6/sqrt(x) = 6/sqrt(4) = 6/2 = 3#

So #x=4# is not a solution of the original equation.

If #x = 9/4# then:

#2sqrt(x) + 1 = 2sqrt(9/4) + 1 = 2*3/2+1 = 3 + 1 = 4#

and #6/sqrt(x) = 6/(sqrt(9/4)) = 6/(3/2) = 12/3 = 4#

So the solution of the original equation is #x = 9/4#