Alpha Beta roots of a Quadratic equation

I'm very sorry, but my textbook doesn't cover this, and when I googled it I'm getting some simple explanations but nothing super helpful. My notes from when I did this over two years ago aren't very clear, and none of my friend's have notes any clearer from when I asked them. I was wondering if someone can help me elucidate them.

again, I'm really, REALLY sorry for having to do this, but asking on here seems to be my last resort.

Lets say that a quadratic equation has roots

\alpha^2

and

\beta^2

.

The product of the roots will be

\alpha^2\beta^2

.
Would that mean that the product is

(\alpha \beta )^2

?

The sum would then be

\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta

?

Are those correct?

Also, I'm clueless about what to do if

\alpha^3

and

\beta^3

come as roots. I know the product will be

(\alpha\ +\beta)^3

but the sum comes out as something seemingly unusable.

for example I say

(\alpha + \beta)^3= \alpha^3 + \beta^3 + 3\alpha^2\beta+3\alpha\beta^2+2\alpha\beta

Soo would that make

\alpha ^3 + \beta^3= (\alpha + \beta)^3-\alpha\beta(3(\alpha+\beta)+2)

???

What were the general equations used for these? And are there any more of these equations I need to learn in order to tackle these questions?

Reply 1

olipal

1

The top two points seem fine, no problems there. Not sure entirely what context this is being used in - what level are you studying at? Your working for the (alpha + beta)^3 seems to be correct as well, if that's all you are trying to show.
EDIT: Sorry, I was pretty tired when I wrote that. As glutamic acid says there was a problem with your expansion of

(\alpha+\beta)^3

Reply 2

GdaT

OP

3

A-levels.

This is mainly the sum and product roots of equations but honestly, we only skimmed it.

Reply 3

Glutamic Acid

17

(\alpha + \beta)^3 = \alpha^3 + 3 \alpha^2 \beta + 3 \alpha \beta^2 + \beta^3 \implies \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3 \alpha \beta ( \alpha + \beta)

, but otherwise you have the right idea.

Reply 4

GdaT

OP

3

Ooooh. I totally multiplied that out wrong. Thanks for that. But it's good to see I have the right idea now.

Well, I know how to use

\frac{1}{\alpha}

and

\frac{1}{\beta}

as roots. I am honestly not sure if there are other variations that I should be aware of.

Now I am looking at my syllabus and I see that we need to know it for cubic equations as well! Our teacher NEVER went over cubics! T_T And it's not in our god forsaken textbook T_T

Well I know for quadratics

\alpha+\beta= \frac{-b}{a}

and

\alpha\beta= \frac{c}{a}

so that's

ax^2-a(\alpha+\beta)+a(\alpha\beta)

But how does this work out for a cubic?

Reply 5

Glutamic Acid

17

Let's suppose our cubic equation is

ax^3 + bx^2 + cx + d = 0

, and writing the roots as

(x-\alpha)(x - \beta)(x - \gamma) = 0

we get the desired relations:

Spoiler

Indeed this generalizes to polynomials of any degree, and are known as Viète's formulas.

Reply 6

GdaT

OP

3

I have to thank you soooo much. I can't even properly express my gratitude. I doubt they'd be cruel enough to give us cubed roots to cubic equations either, but I'll try them out myself later.

Also, thanks for the article. I'll study it in more detail.

Again, big thanks. This was a topic we really didn't go into at all and I think I'm getting it better now.

Reply 7

Pheylan

well i know what i'll be doing at school today

Reply 8

danieltan1995

1

If \alpha, \beta are the roots of 2x^2 + 3x – 4 = 0, calculate the numerical values of \alpha^3-\beta^3? Can someone help me with this?

Reply 9

firingsquad

1

danieltan1995

If \alpha, \beta are the roots of 2x^2 + 3x – 4 = 0, calculate the numerical values of \alpha^3-\beta^3? Can someone help me with this?

have yound found the roots of the quadratic firstly?

Reply 10

Lexie2812

1

helppp whats alpha^3 -beta^3??

Reply 11

16Characters....

13

Original post by Lexie2812

helppp whats alpha^3 -beta^3??

Start with

(\alpha - \beta)^3

and expand and rearrange for

\alpha^3 - \beta^3

Reply 12

B_9710

18

Which exam board are you doing. I know that for AQA in FP1 they only ask about roots related to coefficients of quadratics but in FP2 they ask about cubics aswell.

Reply 13

0le

21

Just like to add, you should never feel ashamed or shy about asking questions. Hope you got the help you need. Come back soon if you do get stuck!

Reply 14

Michael Boom

6

Hello guys, I just started doing Further Maths and I got a homework from page 10, exercise 1C from the book FP1, it would be really helpful if you could help me out with these as I am struggling very hard:
given the roots of equation 3x^2 - 6x +1 =0 are \alpha , \beta , find equation with integer coefficients whose roots are \alpha*\beta^2 and \alpha^2 *\beta.
Please guys help me out, I need to understand how to do those :frown:

(

Reply 15

B_9710

18

Original post by Michael Boom

Hello guys, I just started doing Further Maths and I got a homework from page 10, exercise 1C from the book FP1, it would be really helpful if you could help me out with these as I am struggling very hard:
given the roots of equation 3x^2 - 6x +1 =0 are \alpha , \beta , find equation with integer coefficients whose roots are \alpha*\beta^2 and \alpha^2 *\beta.
Please guys help me out, I need to understand how to do those :frown:

(

The sum of the new roots is

\alpha \beta (\alpha +\beta )

and the product of the new roots is

\alpha ^3 \beta ^3

.
You already know the values of

\alpha + \beta

and

\alpha \beta

.
I see you've digged out quite an old thread there.

(edited 7 years ago)

Reply 16

Prince David

1

given the equation 2x^2 3x-4=0, when required to find a^3-b^3;
a b=-3/2
ab=-2
a^3-b^3=(a-b)[(a b)^2-ab]
but (a-b) cannot be found directly, we therefore make use of the relationship: (a-b)^2=(a b)^2-4ab
=41/4
a-b is therefore = /- (rt41)/2
a^3-b^3 =17rt41/8

Reply 17

Michael Boom

6

Original post by B_9710

The sum of the new roots is

\alpha \beta (\alpha +\beta )

and the product of the new roots is

\alpha ^3 \beta ^3

.
You already know the values of

\alpha + \beta

and

\alpha \beta

.
I see you've digged out quite an old thread there.

Yeah, but there is still a problem with me. I kind of get how you did this one, but the next questions are a bit different and I don't quite get how to do them :frown:

(( I am so frustrated that I don't get this.

Ex:
x^2+4x-6=0 with roots ALPHA, Beta. Find equation for roots ALPHA^2 + BETA and BETA^2 + ALPHA. I feel like i'm so lost when the roots change from beta^2 or alpha^3 or when is 1/(over)b^2 etc...

Reply 18

Michael Boom

6

Original post by B_9710

The sum of the new roots is

\alpha \beta (\alpha +\beta )

and the product of the new roots is

\alpha ^3 \beta ^3

.
You already know the values of

\alpha + \beta

and

\alpha \beta

.
I see you've digged out quite an old thread there.

Thank you very much for the answer but, I don't really know how to use those roots to create the equation, I tried but the result in the back of the book is way different to what I was getting when trying to plug the roots into equation. Please help me :frown:

(edited 7 years ago)

Reply 19

Absolute Madman

Women prefer Alpha roots, if you're a Beta root you're invisible.

Alpha Beta roots of a Quadratic equation

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