How do you solve #4x - (x * 3^(1/2)) = 6#?

1 Answer
Oct 13, 2015

Answer:

#x = 6/13 * (4 + sqrt(3))#

Explanation:

The idea hee is to isolate #x# on one side of the equation.

To do that, start by using #x# as a common factor for the expression that's on the left-hand side of the equation

#4x - x * 3^(1/2) = 6#

#x * (4 - 3^(1/2)) = 6#

You can rewrite the equation by replacing the fractional exponent by its corresponding radical term

#x * (4 -sqrt(3)) = 6#

DIvide both sides of the equation by #(4 - sqrt(3))# to isolate #x# on the left-hand side of the equation

#( x* color(red)(cancel(color(black)(4-sqrt(3)))))/(color(red)(cancel(color(black)(4-sqrt(3))))) = 6/(4 -sqrt(3))#

#x = 6/(4 - sqrt(3))#

Rationalize the denominator of the fraction by multiplying it by #1 = (4 + sqrt(3))/(4 + sqrt(3))#. The expression #(4 + sqrt(3))# is the conjugate of #(4 - sqrt(3))#. The fraction will thus be equivalent to

#6/(4 - sqrt(3)) * (4 + sqrt(3))/(4 + sqrt(3)) = (6 * (4 + sqrt(3)))/((4 - sqrt(3))(4 + sqrt(3))#

#= (6 * (4 + sqrt(3)))/(4^2 - (sqrt(3))^2)#

#= 6/13 * (4 + sqrt(3))#

Therefore, #x# will be equal to

#x = color(green)(6/13 * (4 + sqrt(3)))#