How does the discriminant work (quadratics)

  • Thread starter autodidude
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  • #1
autodidude
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I remember back when I did a lot of quadratic equations, occasionally I had to find an unknown coefficient and if I remember correctly, generally I had to use the discriminant which gave me another quadratic which I could then use to solve for the coefficient.

My question is why does this work? Why can the discriminant be used in this way?
 

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  • #2
chiro
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Hey autodidude.

The discriminant in a quadratic well tell you whether the quadratic has two real roots (discriminant > 0), one real root (discriminant = 0) or no real roots (two complex roots with discriminant < 0).

The discriminant is a nice way to show the above and in certain applications, the nature of the above affects results that build on the result of solving these including things in calculus like solving linear ordinary differential equations.

Basically you can show by completing the square that the solution can be found in terms of the discriminant, but the discriminant is usually just an explicit expression that can be evaluated with a set of given constants. This is the core of why the quadratic equation works.

I get a feeling though you are not talking about an ordinary quadratic in the "special occasions". Maybe you could outline the sort of problem you are referring to if the above is not the case?
 
  • #3
uart
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I think the OP is referring to questions involving determining the range of a parameter in one more more of the coefficients that result in a particular outcome for the number of real roots.

Something for example like: Find the range of parameter "k" such that the quadratic, [itex]x^2 + (2-k)x + 4[/itex], is positive definite.

Is that the type of problem you were thinking of autodidude?
 
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  • #4
chiro
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Ohh ok.

Well in that case it's just adding an extra constraint which means you find that constraint to find your original problem.

Think of it like a tree where the leaves of the tree are all the conditions that you have and you work all the way back to the root to get a final solution.

It's more or less just a means to an end.
 
  • #5
autodidude
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I think the OP is referring to questions involving determining the range of a parameter in one more more of the coefficients that result in a particular outcome for the number of real roots.

Something for example like: Find the range of parameter "k" such that the quadratic, [itex]x^2 + (2-k)x + 4[/itex], is positive definite.

Is that the type of problem you were thinking of autodidude?

Yep, thanks for posting that!

@chiro: Thanks...that makes sense. I think I was expecting some deeper mathematical result lol
 

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