If xy+yz+zx=12-x^2=15-y^2=20-z^2xy+yz+zx=12−x2=15−y2=20−z2 then x+y+z=?x+y+z=?

2 Answers
Narad T.
Jul 29, 2018

The answer is =+-6=±6

Explanation:

Firstly,

xy+yz+zx=12-x^2xy+yz+zx=12−x2

=>⇒, xy+x^2+yz+zx=12xy+x2+yz+zx=12

=>⇒, x(x+y)+z(x+y)=12x(x+y)+z(x+y)=12

=>⇒, (x+y)(x+z)=12(x+y)(x+z)=12

Therefore,

{(x+y=+-3),(x+z=+-4):}

Similarly,

xy+yz+zx=15-y^2

xy+zx+y^2+yz=15

(x+y)(y+z)=15

Therefore,

{(x+y=+-3),(y+z=+-5):}

And finally,

xy+yz+zx=20-z^2

xy+zx+yz+z^2=20

(y+z)(x+z)=20

Therefore,

{(y+z=+-5),(x+z=+-4):}

So,

x+y+x+z+y+z=(+-3)+(+-4)+(+-5)=+-12

2(x+y+z)=+-12

x+y+z=+-12/2=+-6

Before, I considered that (x, y, z) in NN^3 but (x,y,y) in ZZ^3 is also valid.

P dilip_k
Jul 29, 2018
  • xy+yz+zx=12-x^2

=>xy+yz+zx+x^2=12

=>y(x+z)+x(z+x)=12

=>(x+y)(z+x)=3xx4.....[1]

  • xy+yz+zx=15-y^2

=>xy+zx+yz+y^2=15

=>x(y+z)+y(z+y)=15

=>(x+y)(y+z)=3xx5....[2]

  • xy+yz+zx=20-z^2

=>y(x+z)+z(x+z)=20

=>(y+z)(x+z)=5xx4....[3]

It is obvious from [1], [2] and [3]

((x+y)(y+z)(z+x))^2=3^2xx4^2xx5^2

=>(x+y)(y+z)(z+x)=pm(3xx4xx5).....[4]

By [1] and [4]
y+z=pm5

By [2] and [4]
z+x=pm4

By [3] and [4]
x+y=pm3

Summing up we get

2(x+y+z)=pm12

=>x+y+z=pm6

Related questions
Impact of this question
3238 views around the world
You can reuse this answer
Creative Commons License