Question

If $x,y,z$ are real numbers such that $x+y+z=4$ and $x^2+y^2+z^2=6$, then the number of integral value(s) of $x$ satisfying the equation is

Answer 1

$x+y+z=4\Rightarrow z=4-(x+y)\cdots \left(1\right)$

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

Using equation $\left(1\right)$,
$x^2+y^2+(4-(x+y){)}^2=6\Rightarrow x^2+y^2+16+(x+y{)}^2-8(x+y)=6\Rightarrow 2y^2+(2x-8)y+(2x^2-8x+10)=0$
which is a quadratic equation in $y$.

As $y\in R$, so $\Delta \ge 0$
$\Rightarrow (2x-8{)}^2-4\times 2\times (2x^2-8x+10)\ge 0\Rightarrow x^2+16-8x-2(2x^2-8x+10)\ge 0\Rightarrow 3x^2-8x+4\le 0\Rightarrow (3x-2\left)\right(x-2)\le 0\Rightarrow \frac{2}{3}\le x\le 2$

Possible integral values of $x$ are $1,2$.

Hence, the number of integral values of $x$ is $2$.