Question

If $x$ ,$y$ and $z$ are three real numbers such that $x+y+z=4$ and $x^2+y^2+z^2=6$ , then the complete range of $x$ is

• A. $[\frac{2}{3},2]$
• B. $[\frac{4}{3},2]$
• C. $[\frac{2}{3},5]$
• D. None of the above

Answer 1

Answer: (A)

$[\frac{2}{3},2]$

x,y,z are $3$ real numbers

$x+y+z=4,x^2+y^2+z^2=6$

$z=4-(x+y)$

$\therefore x^2+y^2+{(4-x-y)}^2=6$

$2x^2+2y^2+16-8x-8y+2xy=6$

$y^2+(x-4)y+(x^2-4x+5)=0$

As y is real, $△\ge 0$

${(x-4)}^2-4(x^2-4x+5)\ge 0$

$x^2-8x+16-4x^2+16x-20\ge 0$

$-3x^2+8x-4\ge 0$

$3x^2-8x+4\ge 0$

$(3x-2)(x-2)\le 0$

$\therefore xϵ[\frac{2}{3},2]$