SOLUTION: This problem was already answered on this site, but I'm still confused and need clarification. radical (4x + 1) + 3 = 0 First I get the radical by itself radical (4x + 1) = -3;

Question 88217This question is from textbook Beginning Algebra
: This problem was already answered on this site, but I'm still confused and need clarification.
radical (4x + 1) + 3 = 0 First I get the radical by itself radical (4x + 1) = -3; next square both sides to remove the radical: 4x + 1 = 9; next I get the "x" by itself: 4x = 8; divide by 4 therefore x = 2 which when you put back into the equation leaves radical (4(2) + 1) + 3 = 0; radical (9) + 3 = 0; 3 + 3 does not equal zero. Although both my answer and that of the tutors is that equation has no solution, my answer is 2 and his was 0. What am I doing wrong? This question is from textbook Beginning Algebra

Answer by rapaljer(4671) (Show Source): You can put this solution on YOUR website!
I think the tutor may have been saying the answer was x=2, but since it did not work it was rejected. Therefore, the tutor may have been saying NO SOLUTION, which in math is sometimes represented by the Greek letter "PHI". It looks like a zero that has a slash through it, and it is frequently confused with zero. The answer is NOT 0. The answer is NO SOLUTION.

The reason you have a situation like this is because you squared both sides. This is not a "kosher" thing to do, and sometimes it leads to answers that don't really work. This is why when you square both sides, you MUST go back and check the answers to see if they really work. If they don't work in the original equation, they are called "extraneous roots" and they must be rejected.

R^2
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