Linear Transformations of Roots
Linear Transformation of Roots by relationship between roots and coefficients
- If we have the equation: \(x^2-3x-10=0\) which has the roots \(\alpha\) and \(\beta\), what is the equation of the polynomial with roots \(\alpha-1\) and \(\beta-1\)?
- \(\alpha+\beta=\frac{3}{1}=3\)
- \(\alpha\beta=\frac{c}{a}=-\frac{10}{1}=-10\)
- We know that \(\alpha-1+\beta-1=-\frac{b}{a}\)
- \(\alpha+\beta-2=-\frac{b}{a}\)
- Since \(\alpha+\beta=3\):
- \(-\frac{b}{a}=3-2=1\)
We have the first value!
- \((\alpha-1)(\beta-1)=\frac{c}{a}\)
- \(\alpha\beta-(\alpha+\beta)+1=\frac{c}{a}\)
- \(\frac{c}{a}=-10-3+1=-12\)
- Therefore our new equation is \(x^2+x-12\)
- This is a VERY long winded approach, therefore it would be better to use a substitution.
Linear Transformation of Roots using substitution
- If we have the equation: \(x^2-3x-10=0\) which has the roots \(\alpha\) and \(\beta\), what is the equation of the polynomial with roots \(\alpha-1\) and \(\beta-1\)?
- \(W=x-1\)
- \((W-1)^2-3(W-1)-10=0\)
- \(W^2+2W+1-3W-3-10=0\)
- \(W^2-W-12\)
- New function is: \(x^2-x-12=0\)
Using subtitution on a quartic.
- Original polynomial = \(x^4-3x^3+15x+1=0\)
- \(W=2x+1\)
- \(x=\frac{W-1}{2}\)
- Transformed polynomial = \(\frac{(W-1)^4}{16}-\frac{3(W-1)^3}{8}+\frac{15(W-1)}{2}+1=0\)
- Times all by 16
- Expand out
- To be done later!