11But where's a,b, c in the equations?
if x+y+z = 1 and 2xy - z^2 = 1 solve for x, y ,z.
here x, y, z are real numbers...
-
UP 0 DOWN 0 0 11
11 Answers
3@sambit.
using AM ≥ GM is a way as stated by man111 .
x + y ≥ 2√xy
x + y = 1-z
xy = (1 + z^2)/2.
36but ur question is incomplete.....
you should hv added that x,y and z are positive real numbers...
and not only real numbers...
as A.M-G.M inequality holds only for positive real numbers.
21I have a better solution. In my solution I only need x,y,z to be Reals only.
From 1st eqn.
z=1-x-y
Putting the value of z in 2nd eqn.
2xy-z2=1
→2xy-(1-x-y)2=1
which is equivalent to
1+x2+y2-2x-2y=-1
→(x-1)2+(y-1)2=0
→x=1 and y=1
so, z=-1
(x,y,z)=(1,1,-1)
3@rahulmishra
yeah thanks you pointed it out.
here is another one.
x+y = 1-z
so, x^2 + y^2 + 2xy = 1 + z^2 - 2z
also 2xy = 1 + z^2
subtracting the latter equation form the former yields
x^2 + y^2 = -2z
subtracting - 2xy form both the sides.
(x-y)^2 = -2z - 2xy
or, (x-y)^2 = -2z - 1 - z^2 ( since 2xy = 1 + z^2)
or, (x-y)^2 + (z+1)^2 = 0
so, x = y and z = -1.
1708\hspace{-16}\mathbf{2xy=1+z^2}$\\\\ and $\mathbf{x+y+z=1\Leftrightarrow (x+y)=1-z}$\\\\ $\mathbf{*x^2+y^2\geq 2xy}$\\\\ $\mathbf{x^2+y^2+2xy\geq 4xy}$\\\\ $\mathbf{(x+y)^2\geq 2.(2xy)}$\\\\ $\mathbf{(1-z)^2\geq 2+2z^2\Leftrightarrow 1+z^2-2z\geq 2+2z^2}$\\\\ $\mathbf{z^2+2z+1\leq 0\Leftrightarrow (z+1)^2\leq 0}$\\\\ Means $\mathbf{1+z=0\Leftrightarrow z=-1}$\\\\ $\mathbf{(xy=1)}$ and $\mathbf{x+y=2}$\\\\ Means $\mathbf{x=y=1}$\\\\ So $\mathbf{x=y=1\;,z=-1}$\\\\ Given by hsbhatt Sir.......
1708\hspace{-16}$Here $\mathbf{2xy-z^2=1\Leftrightarrow 2xy=1+z^2>0\forall x\in\mathbb{R}}$\\\\ Means $\mathbf{xy>0\forall x\in\mathbb{R}}$\\\\ Now $\mathbf{x+y+z=1\Leftrightarrow x+y=1-z}$\\\\ $\mathbf{(x+y)^2=1+z^2-2z\Leftrightarrow x^2+y^2+2xy=1+z^2-2z}$\\\\ $\mathbf{x^2+y^2=-2z}$\\\\ Means $\mathbf{z<0}$ and here $\mathbf{x,y\neq 0}$\\\\ So if $\mathbf{z<0\;,}$ Then $\mathbf{x+y=1-z>0}$\\\\ So $\mathbf{x\;,y>0}$\\\\ So we can apply $\mathbf{A.M\geq G.M}$