Let the quintic equation: $$ax^5+bx^4+cx^3+dx^2+ex+f=0\,(a\neq 0)$$ Has discriminant $0$, then we can reduce the equation by 2 appropriate Tschirnhaus transformations as described in here. Then we arrive at: $$w^5+w+t=0$$ But because the original quintic as discriminant $0$ or has repeated roots, the Bring-Jerrard quintic also has repeated roots or: $$3125t^4+256=0$$ Then $t$ is one of these 4 values: $$\frac{2\sqrt{2}(1+i)}{5\sqrt[4]{5}},-\frac{2\sqrt{2}(1+i)}{5\sqrt[4]{5}},\frac{2\sqrt{2}(1-i)}{5\sqrt[4]{5}},-\frac{2\sqrt{2}(1-i)}{5\sqrt[4]{5}}$$ And one of the roots of the Bring-Jerrard quintic has the nice corresponding form: $$-\frac{1+i}{\sqrt{2}\sqrt[4]{5}},\frac{1+i}{\sqrt{2}\sqrt[4]{5}},-\frac{1-i}{\sqrt{2}\sqrt[4]{5}},\frac{1-i}{\sqrt{2}\sqrt[4]{5}}$$ And reverting the transformation to obtain a root of the original quintic is also in radicals.
A nice way to revert the quadratic and quartic transformation without encountering extraneous roots is to set: $$x^2+mx+n-y=\frac{x+\alpha_0}{\alpha_1x^3+\alpha_2x^2+\alpha_3x+\alpha_4}=0\,(1)$$ $$y^4+py^3+qy^2+ry+s-z=\frac{y+\beta_0}{\beta_1x^3+\beta_2x^2+\beta_3x+\beta_4}=0\,(2)$$ Then eliminate the denominator, reduce the degree by the original quintic for $(1)$, by the principal quintic for $(2)$. Equate coefficient, find it and: $$x+\alpha_0=y+\beta_0=0$$ Finally, one can solve for $x,y$ without encountering extraneous roots.
Question:
Is this counted as solving in radicals?
Can one possibly do better than this?
