@@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.
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).
Not enough people learn about numerical methods. rearrange the equation as X = SQRT( SQRT(x+5) - 5 ) to find positive root. Put x=1 in Right side.. get value of X and repeat process twice. You get 2.79 in 5 seconds. To get the neg root, x= minus ( equation). Put -1 in for X in RHS and repeat. Not all equations converge but when they do, is seconds. (and why one might learn any language like Python to write the iteration)
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).
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.
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.
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
because
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
@@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.
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).
This is a complicated solution man.
Not enough people learn about numerical methods. rearrange the equation as X = SQRT( SQRT(x+5) - 5 ) to find positive root. Put x=1 in Right side.. get value of X and repeat process twice. You get 2.79 in 5 seconds. To get the neg root, x= minus ( equation). Put -1 in for X in RHS and repeat. Not all equations converge but when they do, is seconds. (and why one might learn any language like Python to write the iteration)
Learning this as a 14 year old is a good or bad idea?
It's a good idea,,,no problem 😊😊
The fact is that it is a jee mains pyq
A bit late in life but it's possible to catch up, good luck.
@@musicsubicandcebu1774 LOL
If you're Terence Tao maybe. The rest of us learn this at a somewhat more advanced age.
learning anything is great
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).
It’s in my head.
who
Is this a standard method to solve such an equation or just an exotic example which by accident can be solved by this tricks?
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.
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.
so crazy, out my eye,useful t
there's gonna be an x^4 term whicb yields there'll be 4 solutions of x.
Srsly this can be solved in 2 to 3 steps
A big pile of Bull!
😮
Substitution:
y**2=x+5
x**2=y+5
---------------------------
(y-x)(y+x+1)=0
😮