Most US College Students Get This Wrong

[@payoo_2674]

2:00 Why not? x⁴-10x²-x+20 = (x²+Ax+B)(x²+Cx+D) = x⁴+(A+C)x³+(D+AC+B)x²+(AD+BC)x+BD = 0 | A+C=0 | D+AC+B=-10 | AD+BC=-1 | BD=20 A=1; B=-4; C=-1; D=-5 x⁴-10x²-x+20 = (x²+x-4)(x²-x-5) = 0 x²+x-4=0 or x²-x-5=0 x=(-1±√17)/2 x=(1±√21)/2 But x²-5 ≥ 0, so x=(-1-√17)/2 or x=(1+√21)/2

[@smh1566]

because

[@nobetana6548]

Q1: how to get this ? x⁴-10x²-x+20 = (x²+x-4)(x²-x-5) = 0 Q2: why the first x=(-1-√17)/2, why not x=(-1+√17)/2 Thanks in advance

[@payoo_2674]

@@nobetana6548 A1: The equation x⁴+ax³+bx²+cx+d=0 can be represented as the product of two polynomials (x²+Ax+B)(x²+Cx+D)=0. By multiplying the contents of both brackets, we get the equation x⁴+(A+C)x³+(D+AC+B)x²+(AD+BC)x+BD=0. By comparing the coefficients with the original equation, we get the system of equations: | A+C=a | D+AC+B=b | AD+BC=c | BD=d By solving it, we get the values ​​of the coefficients A, B C and D. A2: Watch the video at 1:10 and 9:20. The solution must satisfy the condition x²-5 ≥ 0.

[@tassiedevil2200]

On further reflection, I think it is nicer to observe that the quartic x^4 - 10 x^2 -x + 20 has no cubic term and so should factorise as (x^2+x+p)(x^2-x+q), with p*q=20, p+q=-9, q-p=-1 leading to the pair of quadratics in the video (p=-4 q=-5).

[@olaminiranadetula6683]

This is a complicated solution man.

[@nishasunil207]

Learning this as a 14 year old is a good or bad idea?

[@artgallery1058]

It's a good idea,,,no problem 😊😊

[@kaushiksarmah4722]

The fact is that it is a jee mains pyq

[@musicsubicandcebu1774]

A bit late in life but it's possible to catch up, good luck.

[@pietergeerkens6324]

@@musicsubicandcebu1774 LOL If you're Terence Tao maybe. The rest of us learn this at a somewhat more advanced age.

[@DatBom.]

learning anything is great

[@tassiedevil2200]

At the end, instead of reaching for the calculator to check the roots against the requirement x^2-5 >0, you can just look back to the two quadratics: the first says x^2-5=-x-1 is >0, the second that x^2-5=x is >0, and this makes it clear which root is correct in each case. The other two roots are, of course, the remaining roots of (x^2-5)^2=x+5. Note that all four lie in the region (domain) where the square root in the original form is real (x>-5).

[@RyanLewis-Johnson-wq6xs]

It’s in my head.

[@blught4787]

who

[@tomtke7351]

there's gonna be an x^4 term whicb yields there'll be 4 solutions of x.

[@haraldphotode]

Is this a standard method to solve such an equation or just an exotic example which by accident can be solved by this tricks?

[@marilynman]

This is an exotic example, imagine you have a +7 instead of a +5 inside the root; or just about anything in which you cannot substitute into t^2 and t. Another possibility is to have another equation that doesn't result in a depressed polynomial and you have all elements x^4, x^3, x^2, x, c; it would be a nightmare to compute. But then it's good to remember this is an exam question which has a limited time to solve so the method to solve it must be heuristic and mechanical, unless they are testing for how far can you go.

[@pietergeerkens6324]

Yes, this technique for changing variable to yield a polynomial of lesser degree is very powerful. Though uncommon, it's definitely worth learning and practicing. In this case, though, I found it easier to observe that 0 = (x² - ax + b) (x² +ax + c) = x⁴ + (b + c - a²)x² + a(b-c)x + bc gives, comparing to our quartic, the simultaneous equations -10 = b + c - a² -1 = a (b - c) 20 = bc. I always try this first when no obvious rational root can be guessed; and here it's readily seen that a = 1 b = -5 c = -4 is a solution.

[@FlickFix_101]

Srsly this can be solved in 2 to 3 steps

[@peterotto712]

A big pile of Bull!

[@GrahamWest-s6c]

😮

[@angelishify]

Substitution: y**2=x+5 x**2=y+5 --------------------------- (y-x)(y+x+1)=0

[@GrahamWest-s6c]

😮