Proving the Quadratic Equation

[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

 

[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.

 

[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)/2aCheers

 

[uart]

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

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

 

[eXSBass]

RHS - Edit, right hand side!

 

[eXSBass]

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

 

[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 let's 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.

 

[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.

 

[mathwonk]

or is the usual one better? i get

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