Why
$$\LARGE 1 + 3{x^2} \ne 0\quad \quad \quad \forall x \in \mathbb{R} $$

The answers I've came up with...

1. because there's no real number that when we square it would give negative...
2. Parabola doesn't touch the origin [Points (0,0) ]
3. because solving it we get:
$$\LARGE 1 + 3{x^2} = 0$$
$$\LARGE 3{x^2} = -1$$
$$\LARGE {x^2} = -\frac13 $$
$$\LARGE x = \pm \sqrt{-\frac13 } $$
$$\LARGE x = \pm i\sqrt{\frac13 } $$
$$\LARGE x \neq \pm i\sqrt{\frac13 } \quad \quad \forall x \in \mathbb{R}$$
But none of them were true. and this still's my homework...
why:
$$\LARGE 1 + 3{x^2} \ne 0\quad \quad \quad \forall x \in \mathbb{R} $$

Question

Why 
$$\LARGE 1 + 3{x^2} \ne 0\quad \quad \quad \forall x \in \mathbb{R} $$

The answers I've came up with...

1. because there's no real number that when we square it would give negative...
2. Parabola doesn't touch the origin [Points (0,0) ] 
3. because solving it we get:
$$\LARGE 1 + 3{x^2} = 0$$
$$\LARGE  3{x^2} = -1$$
$$\LARGE  {x^2} = -\frac13 $$
$$\LARGE  x = \pm \sqrt{-\frac13 } $$
$$\LARGE  x = \pm i\sqrt{\frac13 } $$
$$\LARGE  x \neq  \pm i\sqrt{\frac13 } \quad \quad \forall x \in \mathbb{R}$$
But none of them were true. and this still's my homework...
why:
$$\LARGE 1 + 3{x^2} \ne 0\quad \quad \quad \forall x \in \mathbb{R} $$

anonymous · Answer

Its because no matter what real value of x you plug into 1+3x^2, the smallest value you can get is 1.  This is because:$$\forall x\in \mathbb{R}, x^2\ge 0\Longrightarrow 3x^2\ge 0 \Longrightarrow 1+3x^2\ge 1\Longrightarrow 1+3x^2>0$$If this doesnt convice you, then just look at the graph of 1+3x^2, and note that it doesnt cross the x axis, so there is no value of x that will make y = 0.

linna · Answer

But sir, that's what I was trying to tell my professors but he said no... another proof ! .. anyway thank you :)