# Finding a value of an expression

• Feb 26th 2011, 04:54 AM
lanierms
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.
• Feb 27th 2011, 01:40 AM
lanierms
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\$.
• Feb 27th 2011, 07:08 AM
NOX Andrew
\$\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?
• Feb 27th 2011, 05:40 PM
lanierms
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\$