discriminant?

[Vastor]

before getting started... i know how to use discriminant for Quadratic Equation/Function

 

which is when [math] b^2 - 4ac > 0 [/math] so, the root(x) would have 2 intersect point

 

and [math] b^2 - 4ac = 0 [/math] so, the root would have 1/equal intersect point

 

and [math] b^2 - 4ac < 0 [/math] so, there are no root.

 

but, I can't get the relation between those..

 

I mean, why [math] b^2?[/math], why [math]-4?[/math] why a and c?, why > or = or < ???

 

http://www.coolmath....arabolas-01.htm

 

this website give me good concept of understanding the Quadratic Equation/Function in General Form/Vertex Form, but I found none for 'discriminant'

 

would be glad if anyone can help

[hypervalent_iodine]

I am by no means skilled at math, but it sounds as though you want to know how to derive the quadratic equation.

 

I found this site, which goes through it step by step:

 

http://www.purplemath.com/modules/sqrquad2.htm

[DJBruce]

So lets take a quadratic equation:

 

[math]f(x)=ax^{2}+bx+c[/math]

 

So the quadratic formula tells us that the roots for this equation will be given by:

 

[math]\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}[/math]

 

So you'll see that the discriminate is just equal to: the term under the radical in quadratic equation. So lets look at the three cases for the discriminate:

 

-[math]b^{2}-4ac=0[/math]

If the discriminate is 0 then we see we get:

 

[math]\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}=\frac{-b\pm \sqrt{0}}{2a}=\frac{-b\pm 0}{2a}=\frac{-b}{2a}[/math]

 

So we see that there is only one real root!

 

-[math]b^{2}-4ac>0[/math]

If the discriminate is greater than 0 we know then that:

[math]\sqrt{b^{2}-4ac}=M[/math]

Where [math]M[/math] is just some positive real number. Therefore the quadratic equation becomes:

 

[math]\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}=\frac{-b\pm M}{2a}[/math]

 

So we know we have two real roots of this quadratic namely [math]\frac{-b+M}{2a}[/math] and [math]\frac{-b-M}{2a}[/math].

 

-[math]b^{2}-4ac<0[/math]

So since the discriminate is negative we know that when we take the square root of a negative number we will get a non-real number, and so if [math]\sqrt{b^{2}-4ac}[/math] is complex we know that the the quadratic equation will give us non-real solutions, and therefore this quadratic will have 0 real roots.

[DrRocket]

So lets take a quadratic equation:

 

[math]f(x)=ax^{2}+bx+c[/math]

 

So the quadratic formula tells us that the roots for this equation will be given by:

 

[math]\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}[/math]

 

So you'll see that the discriminate is just equal to: the term under the radical in quadratic equation. So lets look at the three cases for the discriminate:

 

-[math]b^{2}-4ac=0[/math]

If the discriminate is 0 then we see we get:

 

[math]\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}=\frac{-b\pm \sqrt{0}}{2a}=\frac{-b\pm 0}{2a}=\frac{-b}{2a}[/math]

 

So we see that there is only one real root!

 

-[math]b^{2}-4ac>0[/math]

If the discriminate is greater than 0 we know then that:

[math]\sqrt{b^{2}-4ac}=M[/math]

Where [math]M[/math] is just some positive real number. Therefore the quadratic equation becomes:

 

[math]\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}=\frac{-b\pm M}{2a}[/math]

 

So we know we have two real roots of this quadratic namely [math]\frac{-b+M}{2a}[/math] and [math]\frac{-b-M}{2a}[/math].

 

-[math]b^{2}-4ac<0[/math]

So since the discriminate is negative we know that when we take the square root of a negative number we will get a non-real number, and so if [math]\sqrt{b^{2}-4ac}[/math] is complex we know that the the quadratic equation will give us non-real solutions, and therefore this quadratic will have 0 real roots.

 

All true.

 

I find it easier to understand the implications of the discriminant by simply deriving the quadratic formula and factoring a general quadratic (nothing deep here)

 

[math]ax^{2}+bx+c = a(x^2 + \frac {b}{a} x + \frac {c}{a})[/math]

 

[math]= a (x^2 + \frac{b}{a} x + \frac {b^2}{4a^2} - \frac {b^2}{4a^2} + \frac {c}{a} )[/math]

 

[math] =a([x+\frac {b}{2a}]^2 - \frac {b^2-4ac}{4a^2})[/math]

 

[math] =a([x+\frac {b}{2a}]^2 - [\sqrt {\frac {b^2-4ac}{4a^2}}]^2)[/math]

 

[math] =a( \{[x+\frac {b}{2a}] + [\sqrt {\frac {b^2-4ac}{4a^2}}]\} \{ [x+\frac {b}{2a}] - [\sqrt {\frac {b^2-4ac}{4a^2}}] \})[/math]

 

[math] =a( \{[x +\frac {b}{2a}] + [\frac {\sqrt {b^2-4ac}}{2a}]\} \{ [x+\frac {b}{2a}] - [\frac {\sqrt{b^2-4ac}}{2a}] \})[/math]

 

[math] =a( \{[x- [\frac {- b - \sqrt {b^2-4ac}}{2a}]\} \{ [x - [\frac { -b + \sqrt{b^2-4ac}}{2a}] \})[/math]

 

The roots are thus seen to be [math] [\frac {- b - \sqrt {b^2-4ac}}{2a}][/math] and [math] [\frac { -b + \sqrt{b^2-4ac}}{2a}] [/math] with nature and multiplicity following from the character of [math] \sqrt {b^2-4ac}[/math] which is called the discriminant.

 

The "quadratic formula" is just a way to avoid going through the tedium of completing the square and factoring quadratic polynomials repeatedly. The heart of the matter is a concrete realization of the fundamental theorem of algebra, realizing the general quadratic as a product of two first-order polynomials (and a constant). This may involve complex numbers and repeated roots, but it all falls out of the factorization.

[Vastor]

[math] =a([x+\frac {b}{2a}]^2 - [\sqrt {\frac {b^2-4ac}{4a^2}}]^2)[/math]

[math] =a( {[x+\frac {b}{2a}] + [\sqrt {\frac {b^2-4ac}{4a^2}}]} { [x+\frac {b}{2a}] - [\sqrt {\frac {b^2-4ac}{4a^2}}] })[/math]

 

how come this happen?

 

[math] =a([x+\frac {b}{2a}]^2 - [\sqrt {\frac {b^2-4ac}{4a^2}}]^2) [/math]

should be [math] =a([x+\frac {b}{2a}] - [\sqrt {\frac {b^2-4ac}{4a^2}}])^2[/math]

 

[math] =a( ([x +\frac {b}{2a}] - [\frac {\sqrt {b^2-4ac}}{2a}])( [x+\frac {b}{2a}] - [\frac {\sqrt{b^2-4ac}}{2a}] ))[/math]

 

doesn't it?

[Vastor]

ignore my last post, i'm not so good in algebra that time, anyway..

 

the discriminant seems to be derived from the quadratic formula, and quadratic formula is derived from general form of quadratic equation by moving the 'x' to LHS and others to RHS

 

so, if the degree of 'x' is different (more than 2, etc), does the discriminant for the equation is different???

 

does it's needed for 'full' equation for each degree of polynomial function to be derived to get the discriminant?

 

such as

 

[math]ax^2 + bx + c[/math] for second degree

 

[math]ax^3 + bx2 + cx + d[/math] for third degree

 

etc or wait, did I'm being ridiculous now?

[DrRocket]

ignore my last post, i'm not so good in algebra that time, anyway..

 

the discriminant seems to be derived from the quadratic formula, and quadratic formula is derived from general form of quadratic equation by moving the 'x' to LHS and others to RHS

 

so, if the degree of 'x' is different (more than 2, etc), does the discriminant for the equation is different???

 

does it's needed for 'full' equation for each degree of polynomial function to be derived to get the discriminant?

 

such as

 

[math]ax^2 + bx + c[/math] for second degree

 

[math]ax^3 + bx2 + cx + d[/math] for third degree

 

etc or wait, did I'm being ridiculous now?

 

 

Beyond the quadratic it gets ugly fast.

 

 

http://en.wikipedia.org/wiki/Discriminant