1. Dec 18, 2005

### eXSBass

Hi! I'm new here so if this has been covered please accept my apologies.
I'm having trouble proving the equation from:
ax²+bx+c=0 to the solutions x=(-b(+/-)√b²-4ac)/2a
I've been told to complete the sqaure of ax²+bx+c, then rerrange to give x, but I just can't get it!
If you could show the steps you took to get there it would be very helpful too.
Any help would be appreciated. Thanks

2. Dec 18, 2005

### matt grime

what do you get after completing the square, assuming that you know how to complete the square? and homework ought to go in the homework section.

3. Dec 18, 2005

### eXSBass

I'm sorry. Its not homework. I just need to know how for an exam. When I complete the square I get:
Well, this is what i've got on my sheet. I just don't understand step three!

ax²+bx+c=0
1 a[x²+bx+c]=0 --> x²+(b/a)x+c/a=0
2 (x+b/2a)²+c/a-b²/4a²=0 --> (x+b/2a)²=b²/4a²-c/a
3 (x+b/2a)=((+/-)√b²-4ac)/2a --> x=-b/2a((+/-)√b²-4ac)/2a
4 x=(-b(+/-)√b²-4ac)/2a

I don't understand where on the sheet it got from step 2 to step 3! That is from
(From step 2): =b²/4a²-c/a to
(From step 3): =((+/-)√b²-4ac)/2a

Cheers

Last edited: Dec 18, 2005
4. Dec 18, 2005

### uart

Just put the RHS over a common denominator of $$4a^2$$ and then take the square-root of both sides.

5. Dec 18, 2005

### eXSBass

RHS - Edit, right hand side!

6. Dec 18, 2005

### eXSBass

I understand it now! Thanks alot! It actually makes alot more sense now! Cheers!

7. Dec 19, 2005

### mathwonk

I never could follow that in high school. Now I am not sure why not, too many formulas, or fractions, i guess.

lets try it with no "a", just x^2 + bx + c. then we have to separate out the first two terms from the last:

x^2 + bx +c, then we fill in whatever makes it a square.

thats it i hate fractions, so lets begin with x^2 + 2bx +c,

then we get x^2 + 2bx +c = x^2 + 2bx + b^2 - b^2 + c

= (x+b)^2 + c - b^2. so now if it all = 0, we get (x+b)^2 = b^2 -c,

so x+b = + or - sqrt(b^2-c), and hence x = -b (+or-) sqrt(b^2-c).

now isn't that easier?

then you can solve the gbeneral case ax^2 + bx + c by changing it nito thius one:

i.e. divide by a, getting x^2 + (b/a)X + c/a. then get a 2 in there in the middle as follows:

x^2 + 2(b/2a)x + c/a. then this =0 if and only if

x = -(b/2a) (+or-) sqrt(b^2/4a^2 - c/a)

= -(b/2a) (+or-) sqrt(b^2/4a^2 - 4ac/4a^2)

= -(b/2a) (+or-) sqrt(b^2 - 4ac)/sqrt(4a^2)

= -(b/2a) (+or-) sqrt(b^2 - 4ac)/2a

= [-b (+or-) sqrt(b^2 - 4ac)]/2a.

what a messy formula, i like mine much better.

8. Dec 19, 2005

### mathwonk

e.g. to solve 4x^2 + 6x -1 = 0, change it to
x^2 +(3/2)x -1/4 = x^2 + 2(3/4)x - 1/4, and the solution by my formula
is x = -(3/4) (+or-) sqrt(9/16 + 1/4) = -(3/4) (+or-)sqrt(13)/2.
is that right? it seems nicer than the usual one to me anyway.
my advice to young learners is to have somewhat less respect for tradition.

Last edited: Dec 19, 2005
9. Dec 19, 2005

### mathwonk

or is the usual one better? i get

x =[-6 (+or-) sqrt(36+16)]/8 i guess thats the same, and fewer fractions too! ok tradition has something going for it.