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400 sats \ 5 replies \ @Scroogey 12h \ parent \ on: Brain Not Braining: Part 4 science

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 \ 2 replies \ @noknees OP 8h

How did you determine the coefficients so arbitrarily? The answer is correct though but I think that's cheating. Using factor theorem is a very bad way to solve this kind of equation. Anyways 200 sats for the good effort and somehow getting the correct answer :)

0 sats \ 1 reply \ @Scroogey 7h

I followed

from the link I posted.

0 sats \ 0 replies \ @noknees OP 7h

that's okay, but the rational root test link shows you solved the equation using a solver which is not okay, had you not posted the link and only written "rational root test", I might not have figured it out, but since you did, it puts a wrong impression

it's okay to use a solver once you understand the method, but citing is not always necessary! (another lesson I learnt the hard way)

but another 200 sats for being honest and using formulas the correct way!

0 sats \ 1 reply \ @Scroogey 12h

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

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

0 sats \ 0 replies \ @noknees OP 8h

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