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.

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- Apr 13th 2008, 08:28 AMchrozer[SOLVED] Extraneous Solutions - Logarithms
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. - Apr 13th 2008, 08:38 AMMathstud28Ok
- Apr 13th 2008, 08:41 AMMoo
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 - Apr 13th 2008, 08:44 AMchrozer
- Apr 13th 2008, 08:49 AMMoo
I edited my post :)

And i continue it.

If , that is to say -> ->

The solutions will be and

is clearly inferior to 5 as -> extraneous solution

For , we can discuss about it... - Apr 13th 2008, 08:57 AMSoroban
Hello, chrozer!

Quote:

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.

- Apr 13th 2008, 08:58 AMMoo
GRRRRRRRRRRR

I made a sign error ! Pardon me... - Apr 13th 2008, 09:02 AMMathstud28Ok
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

- Apr 13th 2008, 09:10 AMchrozer
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? - Apr 13th 2008, 09:59 AMMathstud28Ok
- Apr 13th 2008, 10:06 AMchrozer
- Apr 13th 2008, 10:09 AMMoo
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 ;) - Apr 13th 2008, 10:41 AMchrozer