SOLUTION: solve and determine if any of solutions are extraneous (sq rt 3x-2)+ 3 =4x I square each side and get 3x-2=(4x-3)(4x-3)=16x^2-24x+9 =16x^2-27x+11 now what? can't factor

Algebra ->  Test -> SOLUTION: solve and determine if any of solutions are extraneous (sq rt 3x-2)+ 3 =4x I square each side and get 3x-2=(4x-3)(4x-3)=16x^2-24x+9 =16x^2-27x+11 now what? can't factor       Log On


   

Question 857367: solve and determine if any of solutions are extraneous
(sq rt 3x-2)+ 3 =4x I square each side and get
3x-2=(4x-3)(4x-3)=16x^2-24x+9 =16x^2-27x+11
now what? can't factor to set = to 0
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
3x-2=16x%5E2-24x%2B9
16x%5E2-27x%2B11=0
Can't factor? What do you mean? (Yes, it's a bit obscure)
16x%5E2-27x%2B11=%2816x-11%29%28x-1%29=0
Two possible solutions:
16x-11=0
16x=11
x=11%2F16
and
x-1=0
x=1
Now you need to go back and verify that these are actual solutions.
It's critical in this type of problem.