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15 sats \ 3 replies \ @Scroogey 2h \ on: Brain Not Braining: Part 4 science

I would have squared both sides to

 $x^4 - 10x^2 - x + 20 = 0$

and then followed https://math.stackexchange.com/questions/785/general-formula-for-solving-quartic-degree-4-equations/1219804#1219804 with a=0, b=-10, c=-1, d=20. But the quadratic equation

 $v^2 + (2000-27-72*200)v + (100+12*20)^3 = 0$

has no solution, indicating that I made a mistake, and I grew tired :-)

0 sats \ 2 replies \ @noknees OP 2h

well that formula they showed is actually derived from the original quartic formula after depressing it and going case wise depending on the monic polynomial which is not the case for this one, this one is a quasi-symmetric variant, so it's not likely to give the answer :)

Good try tho! Most people won't reach till here...

0 sats \ 1 reply \ @Scroogey 2h

Oh, this is a better strategy: https://www.mathportal.org/formulas/algebra/solalgebric.php

With

a_1=0, a_2=-10, a_3=-1, a_4=20

the cubic equation becomes

 $y^3+10y^2-80y-801=0$

and with

y_1 = -9

(rational root test) the quadratic equations become

 $z^2+z-5=0$

 $z^2+z-4=0$

 $z^2-z-5=0$

 $z^2-z-4=0$

The solutions are

 $\frac{\pm 1 \pm \sqrt{21}}{2}$

and

 $\frac{\pm 1 \pm \sqrt{17}}{2}$

0 sats \ 0 replies \ @Scroogey 1h

Since squaring both sides introduced false solutions, we have to check the four solutions above by inserting them into the original equation, which shows only two of them a true solutions:

 $\frac{1-\sqrt{21}}{2}$

and

 $\frac{-1+\sqrt{17}}{2}$