# Finding a value of an expression

• Feb 26th 2011, 05:54 AM
lanierms
Finding a value of an expression
$x, y, z$ are all different real numbers.

When they satisfy the equation: $x^2-kyz=y^2-kzx=z^2-kxy$,

find the value of $x+y+z+k$.

I already found the value of $k$ though; it's $-2$.
But I can't seem to go any further from that.
• Feb 27th 2011, 02:40 AM
lanierms
Here's how I got the value of K:

$x^2-kyz=y^2-kzx$

$x^2-y^2=kyz-kzx$

$(x+y)(x-y)=-kz(x-y)$

Since $x\not=y$,

$x+y=-kz$.

If we do the same thing for all three pairs, we get:

$x+y=-kz$, $y+z=-kx$, $z+x=-ky$.

If we add all of them together, we get:

$2(x+y+z)=-k(x+y+z)$

$k=-2$.
• Feb 27th 2011, 08:08 AM
NOX Andrew
$x^2 + 2yz = y^2 + 2xz$

$\implies x^2 - 2xz = y^2 - 2yz$

$\implies x^2 - 2xz + z^2 = y^2 - 2yz + z^2$

$\implies (x - z)^2 = (y - z)^2$

$\implies x - z = y - z$ or $x - z = -(y - z)$

$\implies x = y$ or $x = 2z - y$

Does this help any?
• Feb 27th 2011, 06:40 PM
lanierms
I really have no idea how I can use those to get the answer.

$x\not=y$, because they are all different real numbers.

Also, if I try the second one, I get:

$x=2z-y$

$y^2+2zx=z^2+2xy$

$y^2+2z(2z-y)=z^2+2y(2z-y)$

$\longrightarrow y^2+4z^2-2yz=z^2+4yz-2y^2$

$\longrightarrow 3y^2-6yz+3z^2=0$

$\longrightarrow y^2-2yz+z^2=0$

$\longrightarrow (y-z)^2=0$

$\therefore y=z$

But that shouldn't be correct because as indicated earlier: $x\not=y\not=z$