How do you convert depressed quartic to a quadratic?

While solving a quartic equation of the form
#x^4+bx^3+cx^2+dx+e=0#
we make the substitution
#x=y-b/3#
to get an equation of the form
#y^4+py^2+qy+r=0#
now we need to eliminate the linear term to convert it into a quadratic, so how do we do that?

1 Answer
George C.
Jul 22, 2017

You next find the resolvent cubic.

Explanation:

The substitution you need to get the depressed quartic is actually:

#x = y - b/4#

The next stage is to recognise that when a quartic in the form:

#y^4+py^2+qy+r#

is factored as a product of two monic quadratics, then they must take the form:

#(y^2-hy+j)(y^2+hy+k) = y^4+(j+k-h^2)y^2+h(j-k)y+jk#

Hence:

#{ (j+k = h^2+p), (j-k = q/h), (jk=r) :}#

So we find:

#(h^2+p)^2 = (j+k)^2#

#color(white)((h^2+p)^2) = (j-k)^2+4jk#

#color(white)((h^2+p)^2) = q^2/h^2+4r#

Hence we get a cubic equation in #h^2#.

The #3# roots of the cubic correspond to the three ways of splitting the quartic into two quadratic factors (not counting transposing the two factors).

Choose one of the roots of the cubic as #h^2#, then pick either of the two square roots to get #h#. From #h# you can derive #j, k# and thus the two quadratics to solve.

See https://socratic.org/s/aGAWKC2H for an example.

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