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ROOTS-OF-POLYNOMIAL-EQUATIONS

Page history last edited by Steph Richards 15 years, 9 months ago

LINK TO WJEC FP1 PAST PAPERS

 

WJEC SPECIFICATIONS FOR THIS TOPIC

TOPIC NOTES

The number and nature of the roots of polynomial equations, real and complex roots.

Repeated roots.

Candidates will be expected to know that an  nth degree polynomial equation has n roots.

 

The occurrence of the non-real roots of a polynomial equation with real coefficients in conjugate pairs.

  

Relationships between the roots and coefficients of a polynomial equation.

The degree of the polynomial will not exceed 3.

 

These notes are taken from Wikibooks

ROOTS OF QUADRATICS

If we need to find the roots of a given quadratic function we have two formulae that can help us to find the roots of a quadratic equation.

Let \alpha\, and \beta\, be the roots of ax^2+bx+c=0\,. Then, \alpha + \beta=- \frac{b}{a},\quad \alpha\beta=\frac{c}{a}

 

 

ROOTS OF CUBICS

Roots Of Cubic Equations

If we need to find the roots of a given cubic function we have three formulae that can help us to find the roots of a cubic equation.

Let \alpha, \beta\, and \gamma\, be the roots of ax^3+bx^2+cx+d=0\,. Then, \sum\alpha=- \frac{b}{a},\quad \sum\alpha\beta=\frac{c}{a},\quad \alpha\beta\gamma=-\frac{d}{a}

Where: \sum\alpha=\alpha + \beta + \gamma

And: \sum\alpha\beta=\alpha\beta + \alpha\gamma + \beta\gamma

 

 

Comments (2)

lovemath said

at 6:48 pm on Feb 13, 2009

Could anybody here help with this roots question here http://www.thestudentroom.co.uk/showthread.php?t=814687
Currently nobody on the site has been able to solve it using the method I have specified, I can solve it the other way, but would like to understand both methods.

Steph Richards said

at 6:01 pm on Feb 14, 2009

Tried to register on student room but got to wait.
Not sure why you WANT to do it this way as the question part a LEADS you to use sum product etc.
But never mind! I have tried rearranging before cube both sides but sadly got nowhere too. Perhaps you could follow this up
(u^2/3-5u^1/3)^3 =(-7)^3
I couldnt get it to work but variations on this may!

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