Where Does b2-4*a*c Come From: Solving Basic Second Degree Functions?

[mjolnir80]

where does b² - 4*a*c come from? (for second degree functions)

 

[symbolipoint]

where does b² - 4*a*c come from? (for second degree functions)

That is an expression which is contained in the solution of a quadratic equation (yes, a second degree equation). When we start with a*x^2 + b*x + c = 0, we can perform the process of "Completing the square" in order to find a formula for x. Some textbooks on Intermediate Algebra will show this process, both symbolically and graphically.

 

[danago]

adding to what symbol said, the expression you stated is the argument of the square root function in the solution, which is why the nature of the expression determines the nature of the solutions to a quadratic equation.

 

[Eidos]

How is it related to basis functions though?

 

[HallsofIvy]

The quadratic formula is derived from completing the square.

If ax² + bx+ c= 0, subtract c from both sides to get
ax²+ bx= -c

Now, divide both sides by a:
x²+ (b/a)x= -c/a.

We can complete the square by adding (b/2a)²= b²/4a² to both sides:

x²+ (b/a)x+ b²/4a² = b²/4a²- c/a.

The left sides is a perfect square: (x- b/2a)². You can multiply that out to see that.

To add the fractions on the right side, we need to get a common denominator. One denominator is 4a² and the other is a. To get a common denominator I need to multiply numerator and denominator of the last fraction by 4a: (-c/a)(4a/4a)= -4ac/4a²

That gives, for the right side, (b²- 4ac)/4a².

That's where the "b²- 4ac" comes from.

 

[maze]

Basically the whole goal of quadratic equations is to solve the equation ax^2 + bx + c = 0 for x.

It's kind of a trick question though, since sometimes there are no real solutions, depending on what a,b,c are. For example, if a=0, b=0, and c=1, then the equation is 1=0. whoops. For another example, if a=1, b=0, c=1, then the equation is x^2 + 1 = 0, which can't work (think about it). whoops again.

The value of b^2 + 4*a*c will tell you if the equation has a real solution. If it is positive, there are real solution(s). If it is negative, there are no real solutions. If it is zero, there might be a solution or there might not be.

The way people figured all this out is through a process called completing the square, which you can read about here:
http://en.wikipedia.org/wiki/Completing_the_square

 

[d_leet]

If it is zero, there might be a solution or there might not be.

No, there will be a solution.

 

[Eidos]

For another example, if a=1, b=0, c=1, then the equation is x^2 + 1 = 0, which can't work (think about it). whoops again.

No
x^2 + 1 = 0 certainly has roots, they are just complex*. Also as already pointed out if b^2 - 4ac=0 we will only have one root x=\frac{-b}{2a} meaning that our parabola touches the real line with its apex at the root.

* The roots are x=\pm i where i=\sqrt{-1}

 

[maze]

@d_leet

If it is zero, there might be a solution or there might not be.

No, there will be a solution.

Consider b=0 a=0 c=1.

@eidos

For another example, if a=1, b=0, c=1, then the equation is x^2 + 1 = 0, which can't work (think about it). whoops again.

No
x^2 + 1 = 0 certainly has roots, they are just complex*. Also as already pointed out if b^2 - 4ac=0 we will only have one root x=\frac{-b}{2a} meaning that our parabola touches the real line with its apex at the root.

* The roots are x=\pm i where i=\sqrt{-1}

While this is true, we are all talking about real solutions.

It's kind of a trick question though, since sometimes there are no real solutions, depending on what a,b,c are. For example, if a=0, b=0, and c=1, then the equation is 1=0. whoops. For another example, if a=1, b=0, c=1, then the equation is x^2 + 1 = 0, which can't work (think about it). whoops again.

 

[Crosson]

Consider b=0 a=0 c=1.

That is no longer a second degree polynomial.

 

[maze]

That is no longer a second degree polynomial.

Eh, you can split hairs over definitions like this.Some people may define it like you have, but it is often useful to consider a constant as a degenerate polynomial of degree 2, like where you would consider a line segment a degenerate triangle where the triangle inequality is exactly an equality, or where 2 crossing lines are considered a degenerate hyperbola, and so forth. This way of defining it can be useful since each successive set of higher degree polynomials contains the ones of smaller degree as a subset. You also don't have to worry about piecewise definitions and can simply state it is "ax^2+bx+c for any a,b,c in R", as opposed to "ax^2+bx+c for any a in R/{0}, b,c in R"

More importantly, if you don't allow lower degree polynomials to be "in" your set of polynomials of a given degree, you can run into problems if you try to consider the polynomials as making up various function space. For example, what would be the zero element of a function space of polynomials of a certain degree?

 

[Eidos]

While this is true, we are all talking about real solutions.

Just checking that you weren't messing about when you said that x^2 + 1 =0 'cannot work (think about it)'

 

[n_bourbaki]

Eh, you can split hairs over definitions like this.Some people may define it like you have,

Surely everybody defines it like that: the degree is the largest power of x with non-zero coefficient. I can think of nowhere that uses the notion of bx+c to be degenerate quadratic polynomial apart from this thread.

You also don't have to worry about piecewise definitions and can simply state it is "ax^2+bx+c for any a,b,c in R", as opposed to "ax^2+bx+c for any a in R/{0}, b,c in R"

if a is zero then the quadratic formula, which divides by 2a, is nonsense.

More importantly, if you don't allow lower degree polynomials to be "in" your set of polynomials of a given degree, you can run into problems if you try to consider the polynomials as making up various function space. For example, what would be the zero element of a function space of polynomials of a certain degree?

That isn't important at all. So, they're not a vector space? The set of polynomials of degree at most n is a vector space. The zero polynomial is usually taken to have degree minus infinity in order to make the multiplicative property of degree work nicely.

 

[maze]

if a is zero then the quadratic formula, which divides by 2a, is nonsense.

But isn't that's the point... the quadratic formula only applies if a,b,c obey certain propositions.

That isn't important at all. So, they're not a vector space? The set of polynomials of degree at most n is a vector space. The zero polynomial is usually taken to have degree minus infinity in order to make the multiplicative property of degree work nicely.

If you define it without including degenerate cases, then all polynomials of a given degree becomes a rather useless set, wouldn't you say?

The tradeoff between the two definitions being, on one hand you will have to say "polynomials of degree at most n" whenever you want to talk about them as a vector space, vs on the other hand you will have to say "minimum degree" when you want to talk about degree norms for the space.

 

[n_bourbaki]

If you define it without including degenerate cases, then all polynomials of a given degree becomes a rather useless set, wouldn't you say?

Not in the slightest. But you carry on thinking that ax+b is in the "set of degree two polynomials" if you wish - this will put you at odds with just about everyone else I imagine.