How do you solve and check for extraneous solutions in #sqrt(3x - 2) + 3 =4x#?

Start by isolating the radical term on one side of the equation. This can be done by adding #-3# to both sides

#sqrt(3x-2) + color(red)(cancel(color(black)(3))) - color(red)(cancel(color(black)(3))) = 4x - 3#

#sqrt(3x-2) = 4x-3#

Now, because you can't take the squaree root of a negative number if you're working with real numbers, any solution you will find must satisfy two conditions

#sqrt( underbrace((3x-2))_(color(orange)("must be positive"))) = overbrace(4x-3)^(color(red)("must be positive"))#

• #3x-2>=0 implies x >=2/3#
• #4x-3 >=0 implies x >=3/4#

Overall, you need #x>= 3/4#.

Square both sides of the equation to remove the radical term

#(sqrt(3x-2))^2 = (4x-3)^2#

#3x-2 = 16x^2 - 24x + 9#

Rerarrange to get

#16x^2 - 27x + 11 = 0#

Use the quadratic formula to find the two solutions to this quadratic equation

#x_(1,2) = (-(-27) +- sqrt( (-27)^2 - 4 * 16 * 11))/(2 * 16)#

#x_(1,2) = (27 +- sqrt(25))/32#

#x_(1,2) = (27 +- 5)/32 = {( x_1 = (27 + 5)/32 = 1), (x_2 = (27-5)/32 = 11/16) :}#

Notice that only #x_1# satisfies the condition #x>=3/4#, which means that #x_2 = 11/16# will be an extraneous solution.

This is of course because

#11/16 <12/16 = 3/4#

Therefore, your original equation will only have one solution, #x = color(green)(1)#.

Do a quick check to make sure that the calculations are corret

#sqrt(3 * 1 -2) = 4 * (1)-3#

#sqrt(1) = 1 <=> 1 = 1 color(green)(sqrt())#