Factoring without quadradic formula

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The discussion focuses on factoring the quadratic equation -19x^2 - 20x - 9 = 0 without using the quadratic formula (QF). It highlights that this specific equation does not have rational solutions, making traditional factoring impossible. The conversation emphasizes the importance of understanding the discriminant and recognizing when a quadratic may not intersect the x-axis, indicating negative roots. While the quadratic formula can always be used to find roots, the goal is to apply alternative factoring methods as suggested in linked resources. Ultimately, it reinforces the idea that some quadratics cannot be factored in the conventional sense due to their nature.
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Homework Statement


-19x^2-20x-9=0


The Attempt at a Solution


With QF it looks like this:

x=20+-(sqrt):1084/-38

How would I factor this without the QF?
 
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Tebow15 said:
-19x^2-20x-9=0

With QF it looks like this:

x=20+-(sqrt):1084/-38

No, that's not how it looks like. Check the discriminant (part under the square root) again.

Tebow15 said:
How would I factor this without the QF?

This is one that doesn't have rational solutions so you can't factorize it in the usual sense that you're probably thinking. To factor trinomials like this, take a look at this video on how to do so:

But keep in mind that you can always factorize any quadratic if you know the quadratic formula. If a quadratic has roots a and b, then it can be factorized in the form (x-a)(x-b) so since you know the quadratic formula

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

then your two roots are

x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}

x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}

So you can factorize it as

(x-x_1)(x-x_2)= \left(x-\frac{-b+\sqrt{b^2-4ac}}{2a}\right)\left(x-\frac{-b-\sqrt{b^2-4ac}}{2a}\right)

BUT you only need to keep this in mind that it always works, but is extra effort and if you're asked to factorize in an exam, they don't want you to use the quadratic formula, they want you to use the method shown in the video I posted. Also, remember that sometimes quadratics never cross the x-axis, so in these cases, the discriminant would be negative, and you obviously can't take the square root of a negative value.
 
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