If x, y, z are real and distinct,then u = x^2 | vedclass

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

Question

(A) Given $u = x^2 + 4y^2 + 9z^2 - 6yz - 3zx - 2xy$.Multiply and divide by $2$:$u = \frac{1}{2}(2x^2 + 8y^2 + 18z^2 - 12yz - 6zx - 4xy)$Rearrange the

Vedclass · Accepted Answer

Non-negative

Vedclass · Answer

(A) Given $u = x^2 + 4y^2 + 9z^2 - 6yz - 3zx - 2xy$.Multiply and divide by $2$:$u = \frac{1}{2}(2x^2 + 8y^2 + 18z^2 - 12yz - 6zx - 4xy)$Rearrange the terms to form perfect squares:$u = \frac{1}{2}[(x^2 - 4xy + 4y^2) + (x^2 - 6zx + 9z^2) + (4y^2 - 12yz + 9z^2)]$Simplify the expressions inside the brackets:$u = \frac{1}{2}[(x - 2y)^2 + (x - 3z)^2 + (2y - 3z)^2]$Since $x, y, z$ are real,the sum of squares is always non-negative. Thus,$u \geq 0$.

If $x, y, z$ are real and distinct,then $u = x^2 + 4y^2 + 9z^2 - 6yz - 3zx - 2xy$ is always

Similar Questions

The number of solutions of the equations $x+y+z=12$,$x^2+y^2+z^2=50$,and $x^3+y^3+z^3=216$ is

If $x$ is real and $k = \frac{x^2 - x + 1}{x^2 + x + 1}$,then

The number of ordered pairs $(x, y)$ of positive integers satisfying $2^x + 3^y = 5^{xy}$ is

The probability of selecting integers $a \in [-5, 30]$ such that $x^{2}+2(a+4)x-5a+64 > 0$ for all $x \in \mathbb{R}$ is:

If $\sqrt{2} \sin^2 x + (3\sqrt{2} + 1) \sin x + 3 > 0$ and $x^2 - 7x + 10 < 0$,then $x$ lies in the interval

Vedclass Test Series

Exam Paper Generator

Online Exam Module