If x^{2}+y^{2}+z^{2}=2(x+z-1),then find the v | vedclass

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

Question

(D) Given the equation: $x^{2}+y^{2}+z^{2}=2(x+z-1)$.Expanding the right side: $x^{2}+y^{2}+z^{2}=2x+2z-2$.Rearranging the terms to one side: $x^{2}-2

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) Given the equation: $x^{2}+y^{2}+z^{2}=2(x+z-1)$.Expanding the right side: $x^{2}+y^{2}+z^{2}=2x+2z-2$.Rearranging the terms to one side: $x^{2}-2x+y^{2}+z^{2}-2z+2=0$.We can rewrite this as: $(x^{2}-2x+1) + y^{2} + (z^{2}-2z+1) = 0$.This simplifies to: $(x-1)^{2} + y^{2} + (z-1)^{2} = 0$.Since the sum of squares of real numbers is zero only if each term is zero,we have:$x-1=0 \Rightarrow x=1$$y=0$$z-1=0 \Rightarrow z=1$Now,calculate $x^{3}+y^{3}+z^{3} = (1)^{3} + (0)^{3} + (1)^{3} = 1 + 0 + 1 = 2$.

If $x^{2}+y^{2}+z^{2}=2(x+z-1)$,then find the value of $x^{3}+y^{3}+z^{3}$.

Similar Questions

One root of the equation $3x^2 - 10x + 3 = 0$ is $\frac{1}{3}$. Find the other root.

The equation whose roots are $\frac{1}{3 + \sqrt{2}}$ and $\frac{1}{3 - \sqrt{2}}$ is

If $\alpha, \beta, \gamma, \delta$ are the roots of $x^4 - 100x^3 + 2x^2 + 4x + 10 = 0$,then $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{1}{\delta}$ is equal to:

If $\alpha$ and $\beta$ are the roots of the equation $x^{2}-3 \lambda x+\lambda^{2}=0,$ find $\lambda$ if $\alpha^{2}+\beta^{2}=\frac{7}{4}$

The condition that one root of the equation $ax^2 + bx + c = 0$ is three times the other is

Vedclass Test Series

Exam Paper Generator

Online Exam Module