Solving a quartic equation using an unusual idea | An algebraic challenge

[@MizardXYT]

Easier to do this slightly differently: (x² + a)² - (x + b)² = (x² + a - (x + b))(x² + a + (x + b)) -- difference of squares = (x² - x + a - b)(x² + x + a + b) = x⁴ + (2a - 1)x² - 2bx + (a² - b²) = x⁴ + 26x² - x + 182 -- target equation, solve for a,b: 2a - 1 = 26 ⇒ a = 27/2 -2b = -1 ⇒ b = 1/2 a² - b² = (27/2)² - (1/2)² = (729 - 1)/4 = 182 -- check ok! x⁴ + 26x² - x + 182 = (x² - x + 27/2 - 1/2)(x² + x + 27/2 + 1/2) = (x² - x + 13)(x² + x + 14) If the check fails, you'd have to add a factor k in front of (x + b)², which results in solving a cubic.

[@krozjr5009]

3:35 “therefore the right hand side should also be positive if we’re looking for real solutions”. 10:20 All complex roots. I don’t like that.

[@mahdiali4218]

Thanks syber math 2 I watch your video how I become like you

[@barteqw]

I wonder, how can we prove this equation won't have any real solutions? We can check the values on divisors of 182, but is it any faster method? Using differential? In the case of complex, we can also write x as a+bi and look for real values of a and b, right?

[@duytuanlengo5038]

Easy method is let t=x+13, the equation (x + 13)² - x + 13 = 0 become t² - t + 26 = 0 Or (t - 1/4)² + 26 - 1/4 = 0 Obviously the left side is positive in Real.

[@caspermadlener4191]

x⁴+26x²-x+182≥0, because x²-x+1/4≥0, and the rest is non-negative as well.

[@raymondarata6549]

A quick way is to observe that 3 of the 4 terms are always positive for any real x. The only term that can be negative is -x which will only occur when x>0. The coefficient is so small compared to the other 3 terms that for all real x, P(x) > 0. This, of course, means that there are no real roots.

[@kwamezoe4301]

Hw did u get the 169

[@pujachakraborty5244]

Thank u..very much 🙏🙏🙏🙏

[@ShortsOfSyber]

🙏

[@nicolascamargo8339]

Excelente contenido

[@ShortsOfSyber]

Thanks! 🥰