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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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)^{

Vedclass · Accepted Answer

$\frac{3}{2}$

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

Similar Questions

In a college,$\frac{1}{5}$ of the girls and $\frac{1}{8}$ of the boys took part in a social camp. What part of the total number of students in the college took part in the camp?

If $x^{2}+y^{2}-2x+6y+10=0$,then the value of $(x^{2}+y^{2})$ is

If $a = \sqrt{2} + 1$ and $b = \sqrt{2} - 1,$ then the value of $\frac{1}{a + 1} + \frac{1}{b + 1}$ will be

$\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right) \ldots \left(1-\frac{1}{25}\right)$ is equal to

$2+\sqrt{2}+\frac{1}{2+\sqrt{2}}+\frac{1}{\sqrt{2}-2}=?$

Vedclass Test Series

Exam Paper Generator

Online Exam Module