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Without solving the equation, find the value of the roots

  • Thread starter BOAS
  • Start date
  • #1
555
19

Homework Statement



23 - 5x - 4x2 = 0

find ([itex]\alpha[/itex] - [itex]\beta[/itex])2



Homework Equations



In previous parts of the question i've calculated [itex]\alpha[/itex] + [itex]\beta[/itex], [itex]\alpha[/itex][itex]\beta[/itex], 1/[itex]\alpha[/itex] + 1/[itex]\beta[/itex] and ([itex]\alpha[/itex]+1)([itex]\beta[/itex]+1) but I can't think of any rules I know to help me solve the problem.

The Attempt at a Solution



Expanding ([itex]\alpha[/itex] - [itex]\beta[/itex])2

gives

[itex]\alpha[/itex]2+[itex]\beta[/itex]2 -2[itex]\alpha[/itex][itex]\beta[/itex]

But I don't know what I can do with this info.

I'd appreciate a nudge in the right direction!

Thanks
 

Answers and Replies

  • #2
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
3,750
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You can write your expression fairly simply in terms of [itex] \alpha+\beta [/itex] and [itex] \alpha \beta [/itex] (from which you can get the final answer according to your part 2)
 
  • #3
33,507
5,192

Homework Statement



23 - 5x - 4x2 = 0

find ([itex]\alpha[/itex] - [itex]\beta[/itex])2



Homework Equations



In previous parts of the question i've calculated [itex]\alpha[/itex] + [itex]\beta[/itex], [itex]\alpha[/itex][itex]\beta[/itex], 1/[itex]\alpha[/itex] + 1/[itex]\beta[/itex] and ([itex]\alpha[/itex]+1)([itex]\beta[/itex]+1) but I can't think of any rules I know to help me solve the problem.

The Attempt at a Solution



Expanding ([itex]\alpha[/itex] - [itex]\beta[/itex])2

gives

[itex]\alpha[/itex]2+[itex]\beta[/itex]2 -2[itex]\alpha[/itex][itex]\beta[/itex]

But I don't know what I can do with this info.

I'd appreciate a nudge in the right direction!

Thanks
##(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta##
You said that you have already calculated ##\alpha + \beta ## and ##\alpha\beta ##.
 
  • #4
555
19
##(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta##
You said that you have already calculated ##\alpha + \beta ## and ##\alpha\beta ##.
Gah, I should have seen that.

Thanks!
 

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