A Nice, Quick, and Easy Polynomial System

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[vacuumcarexpo]

Let t=x+y+z, then the given expressions can be written as follows:

x^2-x+t-15=0
y^2-y+t-15=0
z^2-z+t-15=0

Therefore, x, y & z are either of the two solutions of k^2-k+t-15=0.

k=(1±√(61-4t))/2

Let α=(1+√(61-4t))/2 & β=(1-√(61-4t))/2, then

t=x+y+z=3α, 3β, 2α+β or 2β+α

t=3α⇒t=9, -15⇒t=9(∵t>0)⇒x=y=z=3
t=3β⇒t=9, -15⇒t=-15(∵t=9 isn't suitable)⇒x=y=z=-5
isn't suitable)⇒(x, y, z)={1-√14, √14, √14}
isn't suitable)⇒(x, y, z)={1+√14, -√14, -√14}

[morteza]

When the conditions are the same for all equations, it means x=y=z🙂
It also means the equality of part to part.
Of course, for the integer roots

[WEAVER]

The smell of new Sybermath in the morning is so refreshing!

[threstytorres]

Since the three eqns have the same sum, we can conclude that the three variables are equal to each other

x + y + z²= 15
=> x + x + x²= 15
=> x² + 2x - 15= 0

Using the Quadratic Formula
[-2 ± Sqrt(2²-4(1)(-15))]/2
=> -5 and 3

Thus the answer(s) is/are:
(3, 3, 3) and (-5, -5, -5)

[morteza]

I found a geometric method to solve quadratic equations.
ax²+bx+c=0

α=arcsin(2c/ar²)/2
X1=R(sin(α))
X2=R(cos(α))
But it is long.

[אסףחברוני]

what about case 3 when (z+y-1)=0 and (y+x-1)=0

[coreyyanofsky]

i assumed x = y = z and found the two symmetric solutions and then assumed x = z, combined the two resulting equations, and ended up with a quartic that factored as
(x - 3)(x + 5)(x^2 - 14) = 0

wasn't sure how to demonstrate that there were no solutions with x ≠ y ≠ z

[Awesome-ctvr]

What i did was:
x+y+z²=15 and subtract another equation
-y-z-x²=-15
They add up to
x-z+z²-x²=0
Then add -z²+x² to both sides
And get
x-z = x²-z² which happens to be:
x-z = (x-z)(x+z) divide both side by x-z
x+z = 1
Subtitute it in the 2nd equation where we have that expression
1 + y² = 15
y²=14
y=+-sqrt14
And then we can repeat the first step
This time choosing different equations
And we get
x+y=1 and we know y
x+sqrt14=1
x=1-sqrt14
Put the new x in x+z=1
And get that z = sqrt 14 aswell
x, y=z ---> 1-sqrt14, sqrt 14, sqrt 14.
Or if we used the negative result in the first place we wouldve gotten the second variation of this

[devondevon]

x =3 y =3 and z =3
x+ z + y^2 = x+ y +z^2 since both = 15
z+ y^2 = y + z^2
z-y = z^2 - y^2
z-y = (z+y)(z-y)
z+y=1 or z-y=0
z-y doesn't = 1, but the value areas 0 then z and y are equal
hence
y+ z +x^2 = x+y+ z^2
z+x^2 = x + z^2
z-x = z^2-x^2
z-x= (z-x) (z+x)
z+x = 1 or z-x =0
here again z-x = 0 hence z and x are equal
hence x= z = y
hence
using x+z + y^2 =15 yields
y+ y + y^2 =15
2y + y^2 =15
y^2 + 2y -15 =0
(y-3)(y+5) = 0
y= 3 and y =-5
y =3 x =3 and z =3 answer

[ZipplyZane]

Shouldn't you also cover case 3, where x+y=1 and y+z=1, if only to show why this can't produce a solution?

[valdemie]

Didn't you miss case 3 where y+z=1 AND x+y=1 ?

[vacuumcarexpo]

My comment was blocked for some reason.

I'll rewrite it.

Let t=x+y+z, then the given expressions can be written as follows:

x^2-x+t-15=0
y^2-y+t-15=0
z^2-z+t-15=0

Therefore, x, y & z are either of the two solutions of k^2-k+t-15=0.

k=(1±√(61-4t))/2

Let α=(1+√(61-4t))/2 & β=(1-√(61-4t))/2, then

t=x+y+z=3α, 3β, 2α+β or 2β+α

t=3α⇒t=9, -15⇒t=9(∵t>0)⇒x=y=z=3
t=3β⇒t=9, -15⇒t=-15(∵t=9 isn't suitable)⇒x=y=z=-5
isn't suitable)⇒(x, y, z)={1-√14, √14, √14}
isn't suitable)⇒(x, y, z)={1+√14, -√14, -√14}

[RahulKumar-ckhr]

Since the three equations are symmetric this we can say that x=y=z
Now by substituting all variables to x in eqn 1 we get x+x+x²=15
So 2x + x ² = 15
Thus x(2+x) =15
There are two possible values for this equation which is x=3 or x=-5

[gopalsamykannan]

x=3;y=3;z=3 also satisfy the equations.

[rakenzarnsworld]

x = 3, y = 3, z = 3
x = -5, y = -5, z = -5

[shrikrishnagokhale]

By just observation, one can give one of the possible answer and i.e.x=y=z=3

[giuseppemalaguti]

Io ho trovato le soluzioni banali x=y=z che danno (3, 3, 3), (-5, -5, -5) poi forse ce ne sono delle altre, boh

[gressorialNanites]

I only looked at the thumbnail, but it's x=3 y=3 z=3 next!

[MyOneFiftiethOfADollar]

It is too bad the terms nice, quick and easy must appear in the thumbnail to induce viewers to click.

One can make up as many of these type of equations as one wants where it all works out “nicely” 😊

[ngdluu]

you make it complicated, when you find out z=y or z+y=1. For example, we use z=y as (1) and replace it in (2) and (3), we can easily solve this, similarly for z+y=1.