1. Tardigrade
  2. Question
  3. Mathematics
  4. If x,y,z∈ R+ and 16(16 x2 + y2 - 4 x y)=z(.16x+4y-z.), then

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. If $x,y,z\in R^{+}$ and $16\left(16 x^{2} + y^{2} - 4 x y\right)=z\left(\right.16x+4y-z\left.\right),$ then

NTA AbhyasNTA Abhyas 2022

Solution:

$(16 x)^2+(4 y)^2+(z)^2-(16 x)(4 y)-(4 y)(z-(z)(16 x)=0$
$\Rightarrow \frac{1}{2}(16 x-4 y)^2+(4 y-z)^2+(z-16 x)^2$
$=0 \Rightarrow 16 x=4 y=z$