Hello mathstud,
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
Hello, chrozer!
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.
What I was doing was not showing you the extraneous answer in this problem but what would make an extraneous answer...because if we got a number x such that x-5<0 then we would have a problem because we would have lets say x=-3 then we would have since its well known that
Oh ok. I didn't know that .
Why did you get rid of the in the equation in the radical part?
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?
Because he wants to compare it to 5, as the equation is defined if and only if x>5.
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