Viet formulas

[fleet tree]

For what values of parameter a is x²-15/4x+a³=0
X²-15/4x+a³=(15/4)²

[fleet tree]

And it need's to be done through vieta's formulas

[brave dagger]

Can u post the original exercise

[fervent sparrow]

Are these two different questions...?

For the first you just need to ensure a³ doesnt make the denominator zero

Try working for what values of x make the fraction zero and go on from there

[fleet tree]

They are not different questions

[fervent sparrow]

Is that fraction supposed to be ewual to 0 or that square?

[fleet tree]

The exercise says

[gusty fog]

Strange. My approach would be to calculate the values of a for which discriminant is >=0 in both equations, and find the common interval.

I have no idea why or how you would apply vietta sums here

[fleet tree]

For what values of parametar a x²-15/4x+a³=0 is equal to x²-15/4x+a³=(15/4)²

[gusty fog]

So both equations are supposed to be equal?

Or are they supposed to be satisfied?

By a common a?

[fleet tree]

Satisfied

By a parametar a

[gusty fog]

Right so using the discriminant approach makes the most sense

[fleet tree]

But I dont know is discriminant bigger than a smaller or equal to 0

[gusty fog]

I have no idea why you would be asked to use vietta sums

[gusty fog]

Are the roots allowed to be imaginary

?

[fleet tree]

No, I'm not at that stage of math

[gusty fog]

Then it assumes that D>=0

Because if D could be anything

then a could also be anything

[fervent sparrow]

Oh yeah I forgot vieta sums has barely anything to do with this

[gusty fog]

Because there would always be some solution

So try to form inequalities

And find the common region

It is a strange question however

[fleet tree]

Yes

[gusty fog]

Come back to us after you try the discriminant approach

[fleet tree]

When I tryed viettas formulas I stopped at one thing and I didnt know how to do it later

[fervent sparrow]

What does sum / product of roots even have to do with this question

[gusty fog]

Because it does not make sense here

The two equations have different roots

What would you try to achieve by setting up these equations?

One really tough approach could be somehow finding the transformation that relates the two equations

[fleet tree]

I dont understsnd question

[gusty fog]

Then use that to get equations between 4 different roots

Then solve for a

But that makes no sense to do as a serious approach

You are better off using discriminants

[fervent sparrow]

The question asks: what values of a allow for some x to give the value of the fraction as 0?

What you have to do: ensure denominators dont go to 0 else the math teacher shits his pants

The problem: vietas relations have to do with the sum and product of the solutions of a polynomial, not 'a'

[gusty fog]

What fraction?

Also what denominator?

[fervent sparrow]

x²-15/4x+a³ no?

[fleet tree]

He also didnt say that a is real number so you can do it easier

[gusty fog]

That's a quadratic however, I don't see where the fraction is

[fervent sparrow]

Yeah if things go to real then you'd have to check discriminant and stuff as well

[fleet tree]

X

[fervent sparrow]

Theres a division sign

[fleet tree]

/ is division yes

[gusty fog]

If a is not real then there is simply insufficient information to solve

[gusty fog]

15/4 x?

[fleet tree]

I want to say x

[fervent sparrow]

Oh wait what

[gusty fog]

Why would the denominator go to zero for any x?

[fervent sparrow]

Ohhh

I thought it was (x²-15)/(4x+a³)

Sorryyy

[fleet tree]

So I will try D=>0 and tell you what I got

[gusty fog]

Ahh no problem, I was confused for a sec

[gusty fog]

Yes

[fervent sparrow]

Yeah the only logical thing to do would be to try x is real

[fleet tree]

Ok I find the a of the first one and the secound one

Then did union of both

Is that an amswer

[fervent sparrow]

I mean critically speaking if the quedtion wants both conditios to be satisfied (which it shouldve specified by stating "AND") then you would take an intersection of the soln

[gusty fog]

No, not the union, but the intersection

[gusty fog]

I think that is understood because the parameter "a" is the same in both equations.

[fleet tree]

Okay thank you

[fleet tree]

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