And what about a ?
chrozer : do you have more info about a ?
To get the solutions, the real solutions i mean (we'll assume this all along), it has to be such as x-5>0 that is to say x>5.
Study the roots of the trinomial :
If there are no real solutions.
If the only solution is wich is an extraneous solution as
What kinds of solutions would be extraneous when solving for in this equation:
I got up to , but I don't know what to do next.
Quadratic Formula: .
There are two roots, one of them is: .
The radical is: .
This makes negative, an extraneous root.
. . [We cannot have the log of a negative quantity.]
Hence, the only root is: .
. . The other is always extraneous.
I also asked someone else about this question and he explained it this way:
If the roots of this equation were to be real,
If , the solution wouldn't be real.
Would this be right?
As 4*10^a is always positive, 25+4*10^a is always superior to 25, so sqrt(25+4*10^a) is always superior to sqrt(25)=5.
Hence x1 is always superior to (5+5)/2=5
And x2 is always inferior to (5-5)/2=0
So the first one is a real solution, the second one an extraneous solution