quadratic trick

[hollow tide]

https://www.youtube.com/watch?v=ZBalWWHYFQc so i found this cool way to solve quadractics but i just dosent work with this question, 9x^2 - 39x - 30 = 0. can anyone help?

[exotic hearth]

1. Wait patiently for a helper to come along.
2. Once someone helps you, say thank you and close the thread with:

+close

3. Feel free to nominate the person for helper of the week in #helper-nominations

[pliant briar]

Well, that method only works when the coefficient of x^2 is 1.
And, well... It isn't anything tgat clever. You're better off just using the regular methods.

By the usual methods, I mean one of the following three:

1. Quadratic formula: always works.
2. Vieta's formulas: may work for integer coefficients (if the roots are not integers, it doesn't really help).
3. Completing the square: always works.

[jolly python]

Yeah you can we get teached this method in school factorisation method
Video does'nt say what to do when x^2 have coefficient other than 1
Let me tell you
Standard Quadratic equation is
ax^2+bx+c=0
Step 1
a×c
Then try to create b into a×c form
For eg
9x^2-39x-30=0
9×(-30) =-270
So we will have to form to numbers from bx when we multipy them they make -270
9x^2-45x+6x-30=0
9x(x-5) +6(x-5) =0
(9x+6) (x-5) =0

So x=5
x=(-6) /9

If you have any doubt feel free to ask

[novel fable]

slide and divide is another great method too

just like above you find how to create b from a×c
(-45)+6 = -39
-45(6) = -30
THEN divide by a to both of them
(x - 45/9)(x + 6/9) = 0

so x = 5, -2/3

[novel fable]

at the last step you could simplified and move the denominator to coefficient of x (by multiply the whole equation by that)

[surreal lynx]

at 2:40 there's actually a quick image showing one way to do it:

namely, divide by 9 so that the quadratic becomes monic:
[x^2 - \frac{39}{9} x - \frac{30}{9} = 0]

[golden spindle]

invariance.

[surreal lynx]

this means we want the roots to sum to 39/9 = 13/3 and multiply to - 30/9 = - 10/3

(13/6 - u)(13/6 + u) = -10/3
169/36 - u^2 = - 10/3
u^2 = 169/36 + 10/3
u^2 = 289/36
u = ± 17/6
so the roots are
13/6 ± 17/6 = -2/3, 5

[pliant briar]

Eh... Still don't see the point to this. The usual three methods are enough.

[surreal lynx]

the primary benefit is probably pedagogical in that it's easier to see why the method works (aside from completing the square)

of course you can derive the other formulas, but it can seem a bit "magical" nonetheless

[pliant briar]

Hm... Well, maybe. Still, I think the usual methods are enough, as two of them always work and the remaining one works very quickly in case of integer solutions.

[surreal lynx]

it's about ease of memorization and understanding what you're doing

[pliant briar]

Then why the need to learn more formulas that do the same thing?

[eternal plume]

because this one is more intuitive, probably

[pliant briar]

I disagree, but ok.

I mean, I get it.

[surreal lynx]

disagree that it's more intuitive?

[pliant briar]

But I see no need for it. The usual three ways are more intuitive and are faaaaaaaar more widely used, anyway.

[surreal lynx]

I don't find (\frac{-b \pm \sqrt{b^2 - 4 a c}}{2a}) to be partcularly intuitive

[golden spindle]

invariance.

[surreal lynx]

I mean it becomes more intuitive when you dig into it

or rather you can build an understanding of it

but it's still somewhat less direct than the method in the video

[novel fable]

for $x²+bx+c = 0, m = -b/2, x = m ± \sqrt{m²-c}$ is more intuitive imo

[golden spindle]

r·cisθ = r(cosθ + ιsinθ)

[pliant briar]

Really? It's as intuitive as breathing to me, honestly. I know by heart how to derive it.

Don't see why the quadratic formula isn't intuitive.

[surreal lynx]

there's a difference between knowing how to derive it and it being intuitive

[jolly python]

I think its not that its not intuitive i think other methods also teached for a certain way
Like i was teached the factorisation method to solve quadratic equation and said when fighting competitive exams where the time is key element this will help. I can see that it takes too much less time when practiced

[pliant briar]

Hm... Well, alright. As long as it helps you!

I just don't really see a way for this to be useful to me specifically. Oh well. Glad it helps you!