How do you solve #(8sqrt( 2x-1)) / 3 = x#?

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Answer 1 · 2016-02-16T02:44:40

First, send the 3 to the other side in multiplication.

#8(sqrt(2x - 1)) = 3x#

#sqrt(2x - 1) = 3/8x#

Get rid of the square root by squaring both sides of the equation.

#(sqrt(2x - 1))^2 = (3/8x)^2#

#2x - 1 = 9/64x^2#

#64(2x - 1) = 9x^2#

#128x - 64 = 9x^2#

#-64 = 9(x^2 + 128/9 + m - m)#

#m = (b/2)^2#

#m = ((128/9)/2)^2#

#m = 16384/324#

#-64 = 9(x^2 + 128/9 + 16384/324 - 16384/324)#

#-64 = 9(x^2 + 128/9 + 16384/324) - 147456/324#

#(-64 + 147456/324)/9 = (x + 128/18)^2#

#126720/2916 = (x + 128/18)^2#

#+-sqrt(126720/2915) - 128/18 = x#

#-0.52 ~= x and -13.70 ~= x#

When you plug both these answers into the original equation, they both make the square root negative inside. This means there is no solution

Hopefully this helps!