Rekation Between Coefficients For a Cubic When Roots Form a Geometric Sequence

If the roots of a cubic equation  
\[ax^3+bx^2+cx+d=0\]
  form a geometric progression is there a relationship between the coefficients?
Yes.
Let the roots be  
\[\alpha , \; \beta , \; \gamma\]
  and divide the equation above by  
\[a\]
  to give  
\[x^3 + \frac{b}{a}x^2+ \frac{c}{a}x+ \frac{d}{a}\]
.
We can write
\[(x- \alpha )(x- \beta )(x- \gamma )=x^3+x(- \alpha - \beta - \gamma_)+x^2(\alpha \beta +\alpha \gamma + \beta \gamma ) +(-\alpha \beta \gamma )=0\]

Equating coefficients gives
\[\frac{b}{a}=-(\alpha + \beta +\gamma )\]
  (1)
\[\frac{c}{a}= \alpha \beta + \alpha \gamma + \beta \gamma\]
  (2)
\[\frac{d}{a}=- \alpha \beta \gamma\]
  (3)
Since the roots form a geometric sequence we can write  
\[\alpha = x, \; \beta =xy, \; \gamma =xy^2\]
.
Using these in (1), (2) and (3) gives
\[\frac{b}{a}=-(x+xy+xy^2)=-x(1+y+y^2)\]
  (4)
\[\frac{c}{a}= x^2y +x^2y^2+x^2y^3=x^2y(1+y+y^2) \]
  (5)
\[\frac{d}{a}=- x^3y^3\]
  (6)
(5) divided by (4) gives  
\[\frac{c}{b}=xy \rightarrow (\frac{c}{b})^3=-(xy)^3=\frac{d}{a}\]
.
Hence  
\[ac^3=b^3d\]
.

Add comment

Security code
Refresh