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400 sats \ 5 replies \ @Scroogey 8 Nov \ 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} + 10 y^{2} - 80 y - 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 21}{2}

and

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

0 sats \ 2 replies \ @noknees OP 8 Nov

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 8 Nov

I followed

from the link I posted.

0 sats \ 0 replies \ @noknees OP 8 Nov

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 8 Nov

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 - 21}{2}

and

\frac{- 1 + 17}{2}

0 sats \ 0 replies \ @noknees OP 8 Nov

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