Dumb maths question

[Maximilian]

Can every quadratic equation be solved by using the quadratic formula?

 

[eLiu]

title didn't lie

Anyway, yes. Every quadratic equation can be written a*x^2 + b*x + c = 0, where a,b,c could be complex numbers (p + i*q). The quadratic formula is derived directly from the 'canonical' form for quadratic equations (complete the square), so it always works.

 

[Maximilian]

Oh my fking god... Seriously? Ive been mystified for years by quadratics and all i have to do is plug the damn equation into that formula twice and bam theres the answers!?

Noone has ever shown me that formula before ever. Theres other ways to do quadratics right? Whatever they are they confuse the shit out of me.

I don't care anymore though i have the formula! I have the power!!

 

[Maximilian]

This guy on yahoo answers seems to just pull (x-5) (x-1) out of thin air:

http://answers.yahoo.com/question/index?qid=20090605041121AAcpGTP

Not to knock whatever hes doing because its the right answer but i have no idea in hell what he did. This is what i did with the formula.

y = x^2 – 6x + 5

x = 6 + sqroot(36 – 20) / 2*1
x = 6 + sqroot(16) / 2
x = 5

x = 6 - sqroot(36 – 20) / 2*1
x = 6 - sqroot(16) / 2
x = 1

 

[busydude]

Theres other ways to do quadratics right? Whatever they are they confuse the shit out of me.

There are other ways.. but those are specific to some kinds of equations..

like.. 2x^2+4x+2=0

Where 4x can be split into 2x+2x and the equation can be simplified further to obtain the roots.

The equation in the OP can be applied to any quadratic equation.

 

[crashtestdummy]

Can every quadratic equation be solved by using the quadratic formula?

It will be correct for any second degree polynomial. It will yield a complex number when the curve does not intersect the x axis (i.e. b^2-4ac will be negative), and will yield two numbers when it intersects twice.

 

[busydude]

y = x^2 – 6x + 5

Assuming y=0;

Like I said before..

6x can be split into 1x+5x -> x^2-1x-5x+5 = 0 -> x(x-1)-5(x-1)=0 -> (x-5)(x-1)=0

 

[Oil]

nvm

 

[schneiderguy]

This guy on yahoo answers seems to just pull (x-5) (x-1) out of thin air:

http://answers.yahoo.com/question/index?qid=20090605041121AAcpGTP

Not to knock whatever hes doing because its the right answer but i have no idea in hell what he did. This is what i did with the formula.

y = x^2 – 6x + 5

x = 6 + sqroot(36 – 20) / 2*1
x = 6 + sqroot(16) / 2
x = 5

x = 6 - sqroot(36 – 20) / 2*1
x = 6 - sqroot(16) / 2
x = 1

The quick way to solve some quadratic equations is to find two numbers that multiply to equal the constant and add up to equal the coefficient of the x term.

So in this case, (-5)(-1) = 5, -5 + -1 = -6

So then the factored form is (x-5)(x-1) = 0

 

[rcpratt]

This guy on yahoo answers seems to just pull (x-5) (x-1) out of thin air:

http://answers.yahoo.com/question/index?qid=20090605041121AAcpGTP

Not to knock whatever hes doing because its the right answer but i have no idea in hell what he did. This is what i did with the formula.

y = x^2 – 6x + 5

x = 6 + sqroot(36 – 20) / 2*1
x = 6 + sqroot(16) / 2
x = 5

x = 6 - sqroot(36 – 20) / 2*1
x = 6 - sqroot(16) / 2
x = 1

Why are you even using the quadratic equation for that. That's easily factored.

The factors must multiply to 5 and add to -6. There's only two ways to multiply to 5 -- 5*1 and -5*-1. 5 and 1 add to 6. -5 and -1 add to -6. There's your answer.

 

[Maximilian]

Yeah i dont like factors much...

Plugging it all into a one size fits all formula is easier for me.

In b4 sucks at maths :whiste:

 

[DrPizza]

If you're going to run into a lot of quadratic equations with integer constants and coefficients, you should really learn how to factor - it's often easier.
Example:
x²+24x-81=0
(x+27)(x-3)=0
gives x=-27,3 Do that with the quadratic formula and you're going to need a calculator.

Alternatively, learn the complete the square method for solving the quadratic formula. For the same problem,
x²+24x-81=0
x²+24x = 81
find half of 24, square it, and add it to both sides
x²+24x + 144 = 81 + 144
the left side is a perfect square:
(x+12)² = 225
take the square root of both sides:
x+12 = +/- 15
*note: 15 squared is 225, and -15 squared is 225, hence the +/-
subtract the 12.
15-12 = 3, and -15-12 = -27. The same answers as above.

Once you learn that method, it takes up a lot fewer lines and spaces than I've typed it. The complete the square method is ideal with the coefficient on x² is 1, and the coefficient on the x is even. Or, if the coefficient on the x term is divisible by twice the coefficient on the x² term.

It takes up this much space (I'll use dots for spacing):
x²-6x=11 *note, if it was x²-6x-11=0, I'd simply change the - 11 into =11
(x-3)²=20
x=3 +/- 2sqrt(5)

After doing a few of these, a decent student will eventually be able to just look at x²-6x-11=0 and write down the answer.

 

[xboxist]

Learn how to factor easy problems like that. It is pretty much THE cornerstone for a lot of the stuff you'll do in college algebra classes. It will make the rest of your life in math classes much easier. Quadratic formula is generally reserved for this situation:

[quadratic equation is presented to you]
"Can I factor this?"
[take the 5-10 seconds needed to attempt to factor it out in your head]
[if NO, proceed to durdle around with the quadratic formula]