By solving the given equations, 

$y^2 + yz + z^2 = 49 \dots \dots ..\left(1\right)$, 

$z^2+ zx + x^2 = 19 \dots \dots \dots .\left(2\right)$,

$x^2 + xy + y^2 = 39 \dots \dots \dots .\left(3\right)$

the value of $x$ is:

Given:

The given eequations are:

$y^2 + yz + z^2 = 49 \dots \dots ..\left(1\right)$, 

$z^2+ zx + x^2 = 19 \dots \dots \dots .\left(2\right)$,

$x^2 + xy + y^2 = 39 \dots \dots \dots .\left(3\right)$

By subtracting (2) from (3) we get

$$\begin{gathered} y^2-z^2+xy-yz=20 \\ \Rightarrow \left[\left(y-z\right)\left(y+z\right)\right]+x\left(y-z\right)=20 \left[by a^2-b^2=\left(a+b\right)\left(a-b\right)\right] \\ \Rightarrow \left(y-z\right)\left(x+y+z\right)=20 ................. \left(4\right) \end{gathered}$$

By subtracting (2) from (1) we get

$$\begin{gathered} y^2-x^2-zx+yz=30 \\ \Rightarrow \left[\left(y-x\right)\left(y+x\right)\right]+z\left(y-x\right)=30 \left[by a^2-b^2=\left(a+b\right)\left(a-b\right)\right] \\ \Rightarrow \left(y-x\right)\left(x+y+z\right)=30 ................. \left(5\right) \end{gathered}$$

By subtracting (3) from (1) we get

$$\begin{gathered} z^2-x^2-xy+yz=10 \\ \Rightarrow \left[\left(z-x\right)\left(z+x\right)\right]+y\left(z-x\right)=10 \left[by a^2-b^2=\left(a+b\right)\left(a-b\right)\right] \\ \Rightarrow \left(z-x\right)\left(x+y+z\right)=10 ................. \left(6\right) \end{gathered}$$

By dividing (5) by (6) we get

$$\begin{gathered} \frac{y-x}{z-x}=3 \\ \Rightarrow y-x=3z-3x \\ \Rightarrow y=3z-2x \end{gathered}$$

By substituting $y=3z-2x$ in (3) we get

$$\begin{gathered} x^2 + x\left(3z-2x\right) + {\left(3z-2x\right)}^2 = 39 \\ \Rightarrow x^2+3zx-2x^2+9z^2+4x^2-12zx=39 \\ \Rightarrow 3x^2-9zx+9z^2=39 \\ \Rightarrow x^2-3zx+3z^2=13 ................. \left(7\right) \end{gathered}$$

By substituting $z=mx$ in (7) we get

$$\begin{gathered} x^2-3\left(mx\right)x+3{\left(mx\right)}^2=13 \\ \Rightarrow x^2-3m{x}^2+3m^2{x}^2=13 \\ \Rightarrow x^2\left(1-3m+3m^2\right)=13 ................ \left(8\right) \end{gathered}$$

By substituting $z=mx$ in (2) we get

$$\begin{gathered} {\left(mx\right)}^2+ \left(mx\right)x + x^2 = 19 \\ \Rightarrow m^2{x}^2+m{x}^2+x^2=19 \\ \Rightarrow x^2\left(1+m+m^2\right)=19 ................ \left(9\right) \end{gathered}$$

By dividing (9) by (8) we get

$$\begin{gathered} \frac{1+m+m^2}{1-3m+3m^2}=\frac{19}{13} \\ \Rightarrow 13+13m+13m^2=19-57m+57m^2 \\ \Rightarrow 44m^2-70m+6=0 \\ \Rightarrow 22m^2-35m+3=0 \end{gathered}$$

Using $m=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, we get

$$\begin{gathered} m=\frac{35\pm \sqrt{{35}^2-4\left(22\right)\left(3\right)}}{2\left(22\right)} \\ \Rightarrow m=1.5 and \frac{1}{11} \end{gathered}$$

Substituting $m=1.5$ in (9), we get

$$\begin{gathered} x^2\left[1+1.5+{\left(1.5\right)}^2\right]=19 \\ \Rightarrow 4.75x^2=19 \\ \Rightarrow x^2=4 \\ \Rightarrow x=\pm 2 \end{gathered}$$

Substituting $m=\frac{1}{11}$ in (9), we get

$$\begin{gathered} x^2\left[1+\frac{1}{11}+{\left(\frac{1}{11}\right)}^2\right]=19 \\ \Rightarrow 1.10x^2=19 \\ \Rightarrow x^2=17.27 \\ \Rightarrow x=\pm 4.16 \end{gathered}$$

Hence, option B is correct.