Help with common form

Hey I was wondering if I could have some help with a certain problem that has been bugging me. I would really like an explination than a direct answer, since I want to be able to comprehend the idea and not just get an easy answer. Hope you can help.

ax^2 = x^2 - x

And I need to find for what values of a does the equation have: no solutions, exactly one solution, and two solutions.

So I'm stuck from here on out, I've tried a lot of different things but basically I think I'm starting the problem wrong. First I need to get it in common form, correct? Can I get some help with this part?

Then I can take it from there, I know to get 1 solution a=0, 2 a>0, and none a<0 correct? And I just use the quadratic formula  part to find the (radical right? get my terms mixed up) b^2-4(a)(c).

So basically, help me get start the problem/ get it in common form and I can take it from there. Thanks a lot.

Sure you typed this question correctly? x=0 is always trivially a solution of this problem.

Yes I'm sure I typed it correctly. Am I completely over thinking this and it has one solution when a=0, 2 when a>0, and none when a<0? Mean I need to justify my answers... huh...

mean the exact problem is:

For what values of a does the equation ax^2=x^2-x
A) have no solutions
b) have exactly one solution
c) have two solutions

Justify your answers.


So do I get it in common form (x^2+x+c) first? I have no idea... any help is great thanks.

The easy way would be to take note of >>2's observation (cannot have no solutions!), and divide out that root.
Thus, you get ax = x-1
(a-1)x = -1
x = 1/(1-a)
Thus, a 2nd solution exists iff a =/= 1

If you must put it in common form for this question to make sense to you, then just shove the ax^2 to the other side.
ie: 0 = (1-a)x^2 - x
"a" = 1-a
"b" = -1
"c" = 0

Huh I kind of get it.

Can you explain the dividing out the root part again?
You find the square root of the entire equation? The square root of x=1? Where do you get the a-1 part?

So what did you do for the first step again for common form?
ax^2  =  x^2-x
-x^2 +x  -x^2 +x
ax^2-x^2+x=0

I could alternatively just subtract ax^2 from bot sides to set it to 0=x^2-x-ax^2

I guess if you could describe each step that you did that would be great. I'm just not seeing how you got the answer you did but with a bit more I should see. Thanks for the help though.

I could alternatively just subtract ax^2 from bot sides to set it to 0=x^2-x-ax^2
This. Then you end up with
(1-a)x^2 - x = 0  [divide by x where x =/= 0]
(1-a)x - 1 = 0  [add 1]
(1-a)x = 1  [divide by 1-a]
x = 1/(1-a)

Wait so can you explain how you got the (1-a) part again? I'm writing out the problem and I just can't see what you did there... so if you can show what you did each step to start it off I could get it.

thanks again for doing this.

If you have 0=x^2-x-ax^2, then you can rearrange and get 0=x^2-ax^2-x
yet x^2-ax^2 is the same as (1-a)x^2
thus you get 0=(1-a)x^2-x

ohhhh I finally get it!!!! bingo I wrote it out and x^2 is essentially 1x^2, and that can be combined with ax^2 because of the similar powers!!

Ohhh thanks a ton! I finally get it! So now I know the values of a and c and can input them in the quadratic formula to find the radical and set each to > = or < to 0 and get my answers.

Ah thank you a ton thanks for putting up with me! I'll let you know if I have any other problems!

Wait... ok problem again.

Putting this in the quadratic formula I end up with

(-1)^2-4(1-a)(0)=1

(just need radical to find the possible solutions)

so does that mean that
a=1 1 solution
a>1 2 solutions
a<1 0 solutions?

Any way I can test this? Thanks for the help again.

>>10
Think of it this way, there are always 2 solutions: when a=1 they are both x=0, otherwise one is x=0 the other is x=1/(1-a).

>>11

So am I on the right track using the quadratic formula? Because I'm looking at similar problems to this that I've done before and you use

SQUARE ROOT of (b^2-4(a)(c)) =, >, or < to 0

So for example I put in (-1)^2-4(1-a)(0)=0   to find 1 solution.

then i got 1=0... but I must have done something wrong because I should be getting something like a=0 or a=1 or something.... help with this part? Thanks

\frac{-b \pm \sqrt {b^2-4ac}}{2a}
Plugging in your values a is 1-a, b is -1, c is 0 so:
\frac{1 \pm \sqrt {-1^2-4(1-a)0}}{2(1-a)}
\frac{1 \pm \sqrt {1}}{2(1-a)}
\frac{1 \pm 1}{2(1-a)}
Getting rid of the ±:
\frac{0}{2(1-a)} = 0
\frac{2}{2(1-a)} = \frac{1}{1-a}

>>13

Huh... so what do those two answers (0 and 1/1-a) mean exactly?

Do I use them to find the values that a has to be for the equation to have 0, 1 and 2 solutions?

>>14
Those are the values of x, the equation always has two solutions (sometimes the two are equal).

Oh so are you saying that there's no way I can have just 1 or 0 solutions? Huh... and I would justify that how?

An n-th order polynomial has n solutions.

Oh yeah I remember that part now! That's the most it can have, and it has all 2 since it is x^2. So 1/0 aren't possible.

Ok thanks a ton I'll write all this out and ask a tutor to double check it in case we're on the wrong track but it makes sense now.

Thanks for the help, i really appreciate it.

ax^2 = x^2 - x
(a-1)x^2 + x = 0
x((a-1)x +1) = 0

so x = 0
or (a-1)x+1  = 0

ie. x = 0
or x = 1/(1-a) if a |= 0.