# Thread: Finding a value of an expression

1. ## Finding a value of an expression

$\displaystyle x, y, z$ are all different real numbers.

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

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

I already found the value of $\displaystyle k$ though; it's $\displaystyle -2$.
But I can't seem to go any further from that.

Here's how I got the value of K:

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

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

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

Since $\displaystyle x\not=y$,

$\displaystyle x+y=-kz$.

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

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

If we add all of them together, we get:

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

$\displaystyle k=-2$.

3. $\displaystyle x^2 + 2yz = y^2 + 2xz$

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

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

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

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

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

Does this help any?

4. I really have no idea how I can use those to get the answer.

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

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

$\displaystyle x=2z-y$

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

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

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

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

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

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

$\displaystyle \therefore y=z$

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