Solving A Quartic | Two Methods

[@hookahsupplier.5155]

Awesome!

[@SyberMath]

Thanks!

[@sidimed1956]

Thank you so mcuh Can you do a video explaining to us how to prepare to IMO ❤❤

[@faridganouche2122]

There is no solotion in R

[@NadiehFan]

My approach to factor the left hand side of x⁴ + 4x² + 4x + 7 = 0 into two quadratics by transforming the left hand side into a difference of two squares would be as follows. We can complete x⁴ + 4x² into a square (x² + 2)² by adding 4, but then we also have to substract this again of course, so we get (x² + 2)² − 4 + 4x + 7 = 0 which is (x² + 2)² + 4x + 3 = 0 To turn the left hand side into a difference of two squares we start by writing this as a difference of two terms, that is, as (x² + 2)² − (−4x − 3) = 0 Now, the first term (x + 2)² at the left hand side is already a square, but the second term (−4x − 3) is not yet a square. The difference between two terms will not change if we add the same thing to both terms. So, the idea is to add something to both terms in such a way that the first term will _remain_ a square and the second term will _become_ a square as well. Taking any number k, we can add 2k(x² + 2) + k² = 2kx² + k² + 4k to both terms, since (x² + 2)² + 2k(x² + 2) + k² = (x² + 2 + k)² so the first term will remain a square regardless of the value of k. Adding 2k(x² + 2) + k² = 2kx² + k² + 4k to each of the two terms at the left hand side we have (x² + 2 + k)² − (2kx² − 4x + k² + 4k − 3) = 0 Now, the first term at the left hand side is a square for any value of k, so we are free to choose k in such a way that the second term 2kx² − 4x + k² + 4k − 3, which is a quadratic in x, will also become a perfect square, that is, the square of a linear polynomial in x. A quadratic ax² + bx + c is a perfect square if and only if this quadratic has two identical zeros (that is, a single zero with multiplicity 2) and this is the case if and only if the discriminant b² − 4ac of this quadratic is zero. So, the condition which k must satisfy in order for the the second term 2kx² − 4x + k² + 4k − 3 to become a perfect square is (−4)² − 4·2k·(k² + 4k − 3) = 0 which gives k³ + 4k² − 3k − 2 = 0 The sum of the coefficients of this cubic in k is zero, so k = 1 is a solution of this cubic equation. With k = 1 our quartic equation (x² + 2 + k)² − (2kx² − 4x + k² + 4k − 3) = 0 becomes (x² + 3)² − (2x² − 4x − 2) = 0 or (x² + 3)² − 2(x − 1)² = 0 or (x² + 3)² − (√2·x − √2)² = 0 which is the same result as in the video, but we only had to deal with a single equation in a single variable k to obtain this result. Using the difference of two squares identity the quartic can now be factored as (x² − √2·x + 3 + √2)(x² + √2·x + 3 − √2) = 0 The discriminants of both quadratic factors are negative, so the quartic has no real solutions. Alternatively, we could also have started by rewriting the equation as (x²)² − (−4x² − 4x − 7) = 0 and then add 2kx² + k² to both terms at the left hand side to get (x² + k)² − ((2k − 4)x² − 4x + (k² − 7)) = 0 where k must satisfy k³ − 2k² − 7k + 12 = 0 to make the second term a perfect square. This is of course the same cubic equation as the cubic equation in a in the video.

[@goldfing5898]

I solved it similarly (see above), according to Ferrari. The only major difference is that rather than writing the difference of two squares on one side, e.g. A^2 - B^2 = 0, and then using the 3rd binomic formula to factorize to (A + B)(A - B) = 0, I prefer to write A^2 = B^2 and then taking the square root to A = +-B, which of course leads to the same results. And I had a slightly different cubic resolvent, but that's normal. For instance, Descartes used a similar but linearly transformed cubic resolvent in comparison to Ferrari.

[@scottleung9587]

Nice, although kinda messy!

[@jinishgaming3240]

I think you put the value of c is wrong please check n let me know

[@lawrencejelsma8118]

a^2 - 2c = 7 is wrong!? The solution can be checked with a root finder computer calculator for x^4 + 4x^2 + 4x + 7 = 0 equation. I didn't realize until looking up scientific calculators they don't have abilities to program a software computer calculation, nowadays. I still use my Muller's Method code in Basic programming language in a 1985 still functioning computer calculator to get roots (use Wolfram Alpha nowadays also by just typing in that polynomial and it's roots found).

[@charlziedouglas-mo7uc]

I thought you were going to use Ferrari's method 😅

[@rakenzarnsworld2]

There are no real solutions for this quartic equation.

[@NowInAus]

Cute and method 2 much neater

[@SyberMath]

Thank you! 😊

[@goldfing5898]

x^4 + 4x^2 + 4x + 7 = 0 x^4 = -4x^2 - 4x - 7 Add 2zx^2 + z^2: x^4 + 2zx^2 + z^2 = (2z - 4)x^2 - 4x + (z^2 - 7) Left side is (x^2 + z)^2 Right side is perfect square iff. D = b^2 - 4ac = 0 or b^2 = 4ac with a = 2z - 4 = 2(z - 2) b = -4 c = z^2 - 7 Therefore (-4)^2= 4 * 2*(z - 2) * (z^2 - 7) 16 = 8 * (z - 2) * (z^2 - 7) 2 = (z - 2)(z^2 - 7) 2= z^3 - 2z^2 - 7z + 14 0 = z^3 - 2z^2 - 7z + 12 Cubic resolvent z^3 + 12 = 2z^2 + 7z z^3 + 12 = z*(2z + 7) z = 3 is a solution, because 3^3 + 12 = 27 + 12 = 39 3*(2*3 + 7) = 3*(6 + 7) = 3*13 = 39 ...

[@SyberMath]

Nice