Question

If $2x^2+3y^2+4z^2-\sqrt{6}xy-2\sqrt{3}yz-2\sqrt{2}xz=0$ then find the value of $(2x^2+3y^2+16z^2+2\sqrt{6}xy-8\sqrt{3}yz-8\sqrt{2}xz)$.

• A. 1
• B. 0
• C. 2
• D. 3

Answer 1

The correct option is B 0

$2(2x^2+3y^2+4z^2-\sqrt{6}xy-2\sqrt{3}yz-2\sqrt{2}xz)=0\times 2$
$\Rightarrow 4x^2+6y^2+8z^2-2\sqrt{6}xy-4\sqrt{3}yz-4\sqrt{2}xz=0$
$\Rightarrow 2x^2-2\sqrt{6}xy+3y^2+3y^2-4\sqrt{3}yz+4z^2+4z^2-4\sqrt{2}xz+2x^2=0$
$\Rightarrow (\sqrt{2}x-\sqrt{3}y{)}^2+(\sqrt{3}y-2z{)}^2+(2z-\sqrt{2}x{)}^2=0$
$\sqrt{2}x-\sqrt{3}y=0\Rightarrow \sqrt{2}x=\sqrt{3}y$
$\sqrt{3}y-2z=0\Rightarrow \sqrt{3}y=2z$
$2z-\sqrt{2}x=0\Rightarrow 2z=\sqrt{2}x$
$\sqrt{2}x=\sqrt{3}y=2z$
$2x^2+3y^2+16z^2-2\sqrt{6}xy-8\sqrt{3}yz-8\sqrt{2}xz$
$(\sqrt{2}x=\sqrt{3}y=2z)$
$=(\sqrt{2}x{)}^2+(\sqrt{3}y{)}^2+16z^2-2\times \sqrt{2}x\times \sqrt{3}y-4\times \sqrt{3}y\times 2z-4\times \sqrt{2}x\times 2z$
$=(2z{)}^2+(2z{)}^2+16z^2-2\times 2z\times 2z-4\times 2z\times 2z-4\times 2z\times 2z$
$=4z^2+4z^2+16z^2+8z^2-16z^2-16z^2$
$=32z^2-32z^2$
$=0$
So, the correct answer is option (b).