A Polynomial Equation | Problem 351

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[@ConradoPeter-hl5ij]

Nice video! Thanks

[@aplusbi]

Thank you too!

[@Arycke]

Nice video! Substitution is dope.
One small thing, and it is probably just a me thing, when you say "plus minus" you write minus plus. Which, for the quadratic equation, doesn't make a difference, of course.
But for angle sums, like cos, sec, (a±b) it flips to minus plus (tan and cot have them, too)
But, it was a small detail that did stand out to me.
Not wrong in this case obv, but yeah

[@aplusbi]

Good point!

[@NadiehFan]

@@aplusbi It works both ways of course. The traditional way of writing the identity for e.g. cos(α ± β) would be
cos(α ± β) = cos α·cos β ∓ sin α·sin β
the idea being that the top sign + at the left combines with the top sign − at the right and the bottom sign − at the left combines with the bottom sign + at the right. But you can equally well write
cos(α ∓ β) = cos α·cos β ± sin α·sin β
and the principle of combining signs in equal positions (top or bottom) at both sides remains valid.

[@NadiehFan]

The equation to solve is
z⁴ + z³ = z² − z − 1
This quartic equation is already in a form suitable for solving using Ferari's method: all terms of a lower than the third degree are on the right hand side and all terms of a third or fourth degree are on the left hand side.
Since z³ = 2·(z²)·(¹⁄₂z) is twice the product of z² and ¹⁄₂z we can complete the square at the left hand side by adding (¹⁄₂z)² = ¹⁄₄z² to both sides, which gives
z⁴ + z³ + ¹⁄₄z² = ⁵⁄₄z² − z − 1
or
(z² + ¹⁄₂z)² = ⁵⁄₄z² − z − 1
Next, we can take any number k and add 2k(z² + ¹⁄₂z) + k² = 2kz² + kz + k² to both sides. Since (z² + ¹⁄₂z)² + 2k(z² + ¹⁄₂z) + k² = (z² + ¹⁄₂z + k)² the left hand side will then remain a perfect square regardless of the value of k. Therefore, adding 2k(z² + ¹⁄₂z) + k² = 2kz² + kz + k² to both sides of our equation we get
(z² + ¹⁄₂z + k)² = (2k + ⁵⁄₄)z² + (k − 1)z + (k² − 1)
Since the left hand side is a perfect square regardless of the value of k, we are are now free to choose k in such a way that the quadratic in z at the right hand will also be a perfect square. It is easy to see that for k = 1 both the linear term (k − 1)z and the constant term (k² − 1) at the right hand side vanish, leaving only the quadratic term (2k + ⁵⁄₄)z² which equals ¹³⁄₄z² = (¹⁄₂√13·z)² for k = 1. So, with k = 1 our equation becomes
(z² + ¹⁄₂z + 1)² = ¹³⁄₄z²
or
(z² + ¹⁄₂z + 1)² = (¹⁄₂√13·z)²
Applying the equal squares property A² = B² ⟺ A = B ⋁ A = −B which says that the squares of two quantities are equal if and only if these quantities themselves are either equal or each others opposite this gives
z² + ¹⁄₂z + 1 = ¹⁄₂√13·z ⋁ z² + ¹⁄₂z + 1 = −¹⁄₂√13·z
and therefore
z² + ¹⁄₂(1 − √13)z + 1 = 0 ⋁ z² + ¹⁄₂(1 + √13)z + 1 = 0
Now we only need to solve these two quadratic equations using the quadratic formula to obtain all four roots of the original quartic equation.

[@scottleung9587]

I got the non-real solutions right, so I assume my real solutions are correct too.