Irreducible Quadratic Equations

[CharlieDoggers]

Hey guys

So this is my first time going through quadratic equations, so it's a bit rocky atm.
Right now I'm struggling to understand irreducible quadratic equations. For example 20 (found at the bottom of this post), my understanding is that the bracketed values are as a result of negating the squared symbol from x2 through finding the square root of x2, which results in the formula being x + 5x + 6 (which I hope is the case?). Where I encounter problems is understanding how factoring the LHS will result in (x + 2)(x + 3) = 0. Is this a result of reducing 6 to the potential multiples that will result in 6 other than 1 (if that makes any sense?) or is it related to somehow breaking down the value of 5x into the numbers 2 and 3?

I also included example 21 because of the fact that I have no idea how the values of + 3 and - 8 result when factorised, let alone what has happened to the value of -24 this whole time in the equation's solution.

Once again, any help would be immensely appreciated

 

[JeffM]

You really are confused, but partly because your text is confusing.

First, every quadratic with real roots is reducible to the form

[MATH]a(x - p)(x - q)[/MATH]
where p and q are the roots. (Notice the signs carefully.) So, the term “irreducible” is, in my opinion, quite misleading.

Second, it is true that many quadratics are difficult to factor. It would be a lot more informative to distinguish between easily factorable and not easily factorable quadratics.

Third, example 20 is not an example of what the text is calling irreducible.

All they are doing is factoring the quadratic.

Do you understand now?

 

[HallsofIvy]

Well, "irreducible" is always "irreducible over" some number system. Typically, it is irreducible over the integers or rational numbers. That is what the text is talking about.

 

[Deleted member 4993]

Hey guys

So this is my first time going through quadratic equations, so it's a bit rocky atm.
Right now I'm struggling to understand irreducible quadratic equations. For example 20 (found at the bottom of this post), my understanding is that the bracketed values are as a result of negating the squared symbol from x2 through finding the square root of x2, which results in the formula being x + 5x + 6 (which I hope is the case?). Where I encounter problems is understanding how factoring the LHS will result in (x + 2)(x + 3) = 0. Is this a result of reducing 6 to the potential multiples that will result in 6 other than 1 (if that makes any sense?) or is it related to somehow breaking down the value of 5x into the numbers 2 and 3?

I also included example 21 because of the fact that I have no idea how the values of + 3 and - 8 result when factorised, let alone what has happened to the value of -24 this whole time in the equation's solution.

Once again, any help would be immensely appreciated

View attachment 25354

In addition to response #2, I would add:

Every quadratic equation (a*x^2 + b*x + c) can be plotted (as y = a*x^2 + b*x + c) as a parabola - with vertical axis of symmetry.

When talk about the solution of quadratic equation (roots of quadratic equation), we are talking about the x-intercepts of the parabola (where the curve intersects the x-axis).

Sometimes the curve MAY not intersect x-axis (imaginary intercept) or the intercepts are irrational numbers (e.g. √2 or π, etc). These are the cases of "irreducible" roots.

 

[JeffM]

@HallsofIvy

I grant that you are correct that a proper use of “irreducible” is relative to some number system, but I see no reason to call

[MATH]x^2 - 5 \pi + 6\pi^2 = 0[/MATH]
irreducible in the context of the way “irreducible“ is being used in the text. There it just means “readily factorable.”