If x^{2}+y^{2}+z^{2}=xy+yz+zx and x neq 0,the | vedclass

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(C) Given the equation: $x^{2}+y^{2}+z^{2}=xy+yz+zx$Multiply both sides by $2$: $2x^{2}+2y^{2}+2z^{2}-2xy-2yz-2zx=0$This can be rewritten as: $(x-y)^{

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$\frac{3}{2}$

Vedclass · Answer

(C) Given the equation: $x^{2}+y^{2}+z^{2}=xy+yz+zx$Multiply both sides by $2$: $2x^{2}+2y^{2}+2z^{2}-2xy-2yz-2zx=0$This can be rewritten as: $(x-y)^{2}+(y-z)^{2}+(z-x)^{2}=0$Since the sum of squares is zero,each term must be zero:$x-y=0 \Rightarrow x=y$$y-z=0 \Rightarrow y=z$$z-x=0 \Rightarrow z=x$Thus,$x=y=z$.Substitute $y=x$ and $z=x$ into the expression $\frac{4x+2y-3z}{2x}$:$\frac{4x+2(x)-3(x)}{2x} = \frac{4x+2x-3x}{2x} = \frac{3x}{2x} = \frac{3}{2}$.

If $x^{2}+y^{2}+z^{2}=xy+yz+zx$ and $x \neq 0$,then the value of $\frac{4x+2y-3z}{2x}$ is

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