How to Solve a Challenging Quartic with a Parameter

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[@mcwulf25]

The second method is far more elegant. When delta is a perfect square, it shouts that there's a simpler method! Nicely explained 😊

[@SyberMath]

Cool, thanks!

[@pyrite2060]

Rewrite as a quadratic in a.
Discriminant simplifies to (2x+1)^2, hence,
2x+1>=0
x >= -1/2 and x

[@dmtri1974]

I loved your 1st method!!😄😄😄 a bit out of the box!! Keep doing this.

[@SyberMath]

Yay! Thank you!

[@levskomorovsky1762]

I solved the equation myself in the first way and was pleased that the root of (2x +1)^2 turned out beautifully, but it turned out that the second method is even more beautiful!

[@matthewfeig5624]

Subtract x^2 from each side. Then factor the left hand side using the difference of two squares.
(x^2-a)^2 - x^2 = a+x-x^2 ---> (x^2-a+x)(x^2-a-x) = -1*(x^2-a-x)
The same factor x^2-a-x appears on both sides. So we have factored the quartic into two quadratics.
(x^2-a+x)(x^2-a-x) + (x^2-a-x) = 0 ---> (x^2-a+x+1)(x^2-a-x) = 0

[@yoav613]

Nice and easy,if you want to solve for x just solve for a😃

[@NadiehFan]

Very nice, especially the second method. I solved this using a more traditional approach, the method of undetermined coefficients to factor a quartic. Expanding the given equation we have
x⁴ − 2ax² − x + (a² − a) = 0
There is no term with x³, so if this is to factor into two quadratics we must have
(x² + px + q₁)(x² − px + q₂) = 0
for some p, q₁, q₂. Expanding this we have
x⁴ + (q₁ + q₂ − p²)x² + p(q₂ − q₁)x + q₁q₂ = 0
and equating corresponding coefficients we have
q₁ + q₂ − p² = −2a
p(q₂ − q₁) = −1
q₁q₂ = a(a − 1)
From p(q₂ − q₁) = −1 we may already suspect that we have either p = 1 and q₂ − q₁ = −1 or p = −1 and q₂ − q₁ = 1 and then it is not hard to find the solution triple (p, q₁, q₂) = (1, −(a − 1), −a) as well as (p, q₁, q₂) = (−1, −a, −(a − 1)) because inverting the sign of p will simply swap the values of q₁ and q₂ for the same factorization. With (p, q₁, q₂) = (1, −(a − 1), −a) we have
(x² + x − a + 1)(x² − x − a) = 0
so
x² + x − a + 1 = 0 ∨ x² − x − a = 0
and then it is just a matter of solving these two quadratic equations in x using the quadratic formula.

[@bobbyheffley4955]

If a=1, one of the solutions of the right-hand equation is the golden ratio.

[@SyberMath]

Wow!

[@giuseppemalaguti435]

Posto y=a-x^2....risultano soluzioni per x+y=0....y-x=1....1)x+a-x^2=0....2)a-x^2-x=1

[@Annihilator-01]

Keep working my friend

[@SyberMath]

Thank you, I will

[@ilanbar1970]

I wonder if there is elegant way to graph parametric equations .. maybe some 3d view in this case

[@SyberMath]

There must be

[@nikos4677]

i did the second one with different substitution

[@GirishManjunathMusic]

I worked this out before watching the video:
(a - x²)² = a + x
setting a - x² = y → x² = a - y:
y² = a + x
x² - y² = - (y + x)
x² - x² + x + y = 0
(x + y)(x - y) + (x + y) = 0
(x + y)(x - y + 1) = 0
Case 1 : x + y = 0:
x + a - x² = 0
x² - x - a = 0
x = ½ ± √(¼ + a)
x = ½ ± √(a + ¼)
Case 2: x - y + 1 = 0
x - a + x² + 1 = 0
x² + x - a + 1 = 0
x = -½ ± √(¼ + a - 1)
x = -½ ± √(a - ¾).

[@broytingaravsol]

so easy