A Homemade Polynomial System Solved in Two Ways

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[felaxwindwalker]

I mostly used the second method, with the biggest difference being that I added 4*equation 1 to equation 2 like you did to get x+y=3, and then subtracted 4*equation 1 from equation 2 to get x-y=1. From there it was simple elimination to get the x and y values.

[rajeshbuya]

Interestingly, you chose to use the binomial form and found (x+y)^4 = 81. You could use the same binomial form (with a simple switch of signs) and find (x-y)^4 = 1. Makes it a lot simpler!

[Chrisoikmath_]

3nd method:
xy(x^2+y^2)=10
(x^2+y^2)^2+4(xy)^2=41
Let a=x^2+y^2 and b=xy
Now solve for a and b using the known identidy (x+y)^2=x^2+y^2+2xy
After use Vieta's follulas for S and P to get the same results

[jmart]

Similar to your 2nd method.
You can rewrite the eqs. to
xy(x^2+y^2)=10
(x^2+y^2)+4(xy)^2=41
Then xy=u and x^2+y^2=v
Solve for u and v and then for x and y.
Another interesting problem.

[jfktvrv]

I added 4 of eq1 to eq2 to get 81=((x^2+y^2)+2xy)^2
So then +_3sqrt(+_1)-x=y
I then substituted this into eq1 and found no solutions for Re(eq1)=0 and Im(eq1)=0 for the +_3i-x=y and for +_3-x=y gave (1, 2), (3-i/2, 3+i/2) and then negative and flipped versions

[minhdoantuan]

Let A=x^2 + y^2 and B=2xy, easily checked that A >= B
1st equation: A×B = 20
2nd equation: A^2 + B^2 = 41, so (A+B)^2 = 81, bc A+B = (x+y)^2 >= 0, A+B = 9
A+B =9, AB=20, A and B are solutions to the equation:
X^2 - 9X + 20 =0, X=4 or X=5
A >= B so A= x^2 + y^2 =5, B= 2xy =4, xy=2
A+B = (x+y)^2 =9, x+y = 3 or -3
Respectively found all solution of (x, y).

[paulkolodner]

Here's a much simpler solution, starting with the result that x + y = +3. Write x = 3/2 - t and y = 3/2 + t. Re-write the first equation as x*y*(x^2 + y^2) = 10 and substitute these values of x and y into it. The result is (9/4 - x^2)*(9/4 + x^2) = 5, or 81/16 - x^4 = 5. So x^4 =1/16. No mess, no quadratic or higher-order polynomial.

[scottleung]

I got the positive solutions, but forgot about the negative ones. BTW that second method was ingenious and the graph was really cool!

[kaarthikananthanarayanan]

SIMPLY SUPERB to watch geniuses like you in youtube.

[kaarthikananthanarayanan]

Hey, I SIMPLY love your videos. There's actually a third method where you equate 1st equation is x^2y^2=100/[(u+ 1/u)+2]^2 and similarly the second equation is 41/{[u+ 1/u]^2+ 4} = x^2y^2; where u=x^2//y^2. We put (u+1/u)=w and solve for w, and then get u and proceed.

[danandsaf]

You can make xy=a and x^2+y^2= b
And you will get ab=10 and b^2 +4a^2=41

[j.r.]

You didn't fully commit to your second method. Besides getting (x + y)^4 = 81, you can also easily find that (x - y)^4 = 1. Then you get a system of two very simple linear equations, which quickly provides all four (x, y) solutions.

[timeonly]

Looking thru all the methods by everyone, it's really true that "There is more than one way to skin a cat!" :-)

[ivitta]

x+y=±3, x-y=±1 (from (x-y)⁴=x⁴-4x³y+6x²y²-4xy³+y⁴=41-40=1), so you have 4 linear systems in x and y

[heribertogutierrez]

Excelente procedimiento y razonamiento felicitaciones, gracias por la información

[giuseppemalaguti]

Pongo s=x+y, p=xy, risulta p=2, s=3, quindi (2, 1), (1, 2) e p=5/2, s=3 che danno soluzioni complesse... Dovrebbe essere valida anche la soluzione p=2, s=-3, cioe (-1, -2), (-2, -1)

[Bhaskar]

I have Some Questions and Theoram As You. I do streaming As you But I don't know making Video as you .How to Record Video as you

[Anik_Sine]

Will someone please explain me when can we take y = kx?

[cmagicalex]

you lose 12 solutions with complex number

[neuralwarp]

But what if y ≠ ky ?
What if y = xª for example ?
What if y = x+c ?