Solving A Quartic Equation | x⁴+12x+3=0

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[MichaelRothwell]

My solution is pretty similar to yours, but is a bit simpler as I just have one unknown to solve for instead of three.

I also use Ferrari's method for solving quartics: rewrite the quartic equation as the square of a quadratic equals the square of a linear expression.

Start by rewriting the quartic as
x⁴=-12x-3
Since the quartic is monic and depressed (i.e. has no x³ term), our quadratic is of the form x²+m.
Since (x²+m)²=x⁴+2mx²+m²,
adding 2mx²+m² to each side of our quartic equation we get the equivalent equation
(x²+m)²=2mx²+m²-12x-3
or (x²+m)²=2mx²-12x+m²-3

So we need to find m such that the RHS is a perfect square, i.e. the discriminant Δ=12²-4×2m×(m²-3)=0
144-8m³+24m=0
8m³-24m-144=0
m³-3m-18=0

By inspection, m=3 is a solution
So we get

(x²+3)²-6(x-1)²=0

x²+√6x +3-√6=0 or x²-√6x +3+√6=0

x²+√6x +3-√6=0:
This has the two real roots
x=(-√6±√[6-4(3-√6)]/2
=(-√6±√[4√6-6)]/2

x²-√6x +3+√6=0:
This has the two non-real complex conjugate roots
x=(√6±√[6-4(3+√6)]/2
=(√6±√[-6-4√6)]/2
=(√6±i√[6+4√6)]/2

So the four solutions are
x=(-√6±√[4√6-6)]/2, x=(√6±i√[6+4√6)]/2.

[randomjin]

Let's write it as "quadratic in 0": 1∙0²+(2x²+𝛿)∙0+(x⁴+12x+3) = 0 for some 𝛿. Then D = 4x²𝛿-48x+𝛿²-12. We want to find such 𝛿 that the discriminant becomes a perfect square. And we're in luck, 𝛿=6 is an obvious choice here. We then have D = 6(4x²-8x+4) = 6(2x-2)². Therefore if we write the original equation as 1∙0²+(2x²+6)∙0+(x⁴+12x+3) = 0 then via quadratic formula we have 0 = -(2x²+6)±(2x-2)√6 which in fact yields just two quadratics in x.

[nadeeraudayanga]

The famous Ferrari's formula presented in a different way.!! 😜😜😜

[dentonyoung]

Well, that was an interesting route to get to the solutions. I learn a lot from this channel.

[shashankkatiha]

Add and subtract 9x^2 to the expression to make x^4 - 9x^2 + 9x^2 + 12x -3 =0 and factorize the first two terms and the last three terms separately.

[giuseppemalaguti]

Dopo opportuni passaggi per risolvere le quarti he risulta... (x^2+3)^2-6(x-1)^2=0, molto semplice da risolvere

[kahouaoieong]

b^3-3b=18, should have 3 roots as well. if we use the other roots (a pair of non-real roots) for b, i’m guessing it will still have the same conclusion?

[robyzr]

dear teacher perfect as usual !! I wish the best for this new 2024 ..

[scottleung]

Took me a few hours, but I got the correct solutions!

[alextang]

Is this a "general" way to solve this kind of Quartic Equation? This is the first time I see. Very nice!!! 👍👍👍👍👍👍

[deltalima]

Wth happened to all the other solutions he was gonna check?

[giuseppeimbimbo]

I wrote the following quartic equation:
x^4-16x^2+20x+3=0
x=3 is a solution
Finding the other solutions was a bit tricky!!!
I tried this method but I gave up...
Can you help me please?