$x(x^2-2)=0$, The answers are $x = 0, \sqrt{2}$, how do I get there? $$x(x^2-2)=0$$
The answers are $x=\sqrt{2}, 0$ how do I get there?
 A: First begin with noting that your polynomial is of degree $3$. This should tell you that there are $3$ (possible complex) roots. Given $x(x^2 -2) = 0$, it follows that at least one of these factors has to be zero. So possible values are $x = 0$ and $x^2 - 2 =0$. That gives you one solution, $x=0$, the other two follow from a special factorization: $a^2 - b^2 = (a-b)(a+b)$. Using this on the factor $x^2-2$, this gives us : $x^2 - 2 =(x - \sqrt{2}) (x + \sqrt{2}) = 0$. It follows that $(x - \sqrt{2}) = 0$ or $(x + \sqrt{2}) = 0$. Combining this with our first solution $x = 0$, we find that all roots to this equation are $x= 0 , \pm \sqrt{2}$
A: $1.$ A product is $0$ if and only if one of the factors is $0$.
$2.$ $x^2-2=0\iff x^2=2$, and the latter means by definition of a square root that $x=\sqrt 2$ if $x>0$.  If $x<0$, setting $x=-t,\enspace t>0$, we have $\;x^2=(-t)^2=t^2=2$, hence $t=\sqrt2$ by the previous case, so $x=-\sqrt 2$.
A: 
If $x(x^2-2)=0,$ then either (at least one) of the factors have to be 0.(because something multiplied by zero is only zero)
$\therefore x=0\space \mathrm{or} \space x^2-2=0$
$\implies x=0,\pm\sqrt2$

A: $x(x^2 - 2)=0$
implies $x=0$ or $x^2 - 2=0$
So $x=0$ or $x^2 = 2$.
So $x=0$, $x= \pm \sqrt{2}$
A: First recall some elementary algebra that if $A\cdot B = 0$, then $A = 0$ or $B = 0$ or both of them $A = 0, B = 0$. So $x(x^2-2) = 0 \to x = 0$ or $x^2-2 = 0$, and $x^2-2 = 0 \to (x-\sqrt{2})(x+\sqrt{2}) = 0 \to x-\sqrt{2} = 0$ or $x +\sqrt{2} = 0 \to x = \sqrt{2}$ or $x = -\sqrt{2}$. Thus there are $3$ solutions: $x = 0,-\sqrt{2}, \sqrt{2}$.
A: Let's start from the beginning.
You are given

$$x(x^2 - 2 ) = 0$$

What are they asking for here?
Well, they're asking for any value of $x$ which would cause the given statement to become true.
What does this mean?
It means that if you were to substitute a number in place of $x$, then solve out the equation, the resulting statement must make logical sense.
How do we find these values of $x$?
We can make a few observations that will aide us in the discovery of these values. For instance, we could separate the equation into two parts then solve each individually.
Observe,
$$x(x^2-2) = 0$$
$$x\qquad(x^2-2) = 0$$
$$x=0,\qquad(x^2-2) = 0$$
Why do we do this?
Well, we want to find a value for $x$ that makes the whole equation true, and since the equation is just a product, if a $0$ is the result anywhere in the equation the entire product will become $0$.
Less talk, more proof.
$$
\begin{align}
x(x^2-2) &= 0\qquad x(x^2-2) = 0\\
(0)(x^2-2) &= 0\qquad\qquad\; x(0) = 0\\
0 &= 0\qquad\qquad\quad\ \ \  0 = 0
\end{align}
$$
$$\color{green}{True!}$$
So , we can break up this equation into two parts and just solve each individually for an $x$ that will be equal to $0$.
Now let's start actually doing this.
We have already divided up the equation above
$$x=0,\qquad(x^2-2) = 0$$
As you can see, we already have a value for $x$ given just by dividing up the equation, that value being $x=0$. Now, we have to see if there are any other possible solutions that can be given by solving the other part of the equation.
Let's look at
$$(x^2-2) = 0$$
All we have to do here is a bit of algebra to solve for $x$.
$$
\begin{align}
x^2-2 & = 0\\
x^2-2+2 & = 0+2\\
x^2 & = 2\\
\sqrt{x^2} & = \pm\sqrt{2}\\
x & = \pm\sqrt{2}\\
\end{align}
$$

Why is there a $\pm$ in front of $\sqrt{2}$?
That is a 'plus or minus' sign, it literally means that the following value can be either positive or negative.
Why do we need it?
Well, squaring a real number will  always produce a positive number, regardless of whether the real number being squared is positive or negative.
$$(2)^2 = 4$$
$$(-2)^2 = 4$$
Therefore, when we are undoing a square, in other words taking a square root, we cannot assume that the result is just positive or negative, since it could be either and still be true. So, we have to split the result of a square root up into two components, its positive result and its negative result, and solve the resulting equations using each result in turn.
In summary,
$$
\begin{align}
x &= \pm\sqrt{2}\\
\implies x=\sqrt{2}&,\quad x=-\sqrt{2}\\
\end{align}
$$

Now that we have our values for $x$, we have to check them to make sure nothing was done incorrectly.
$$\color{red}{\text{Always check your work!}}$$
Given:
$$x(x^2 - 2 ) = 0$$
Tests:
$$x=0\qquad x=\pm\sqrt{2}$$
We see:
$$
\begin{align}
x(x^2 - 2 ) &= 0 \qquad\qquad\quad x(x^2 - 2 ) = 0 \qquad\qquad\qquad\ \ \  x(x^2 - 2 ) = 0\\
(0)((0)^2 - 2 ) &= 0 \qquad (\sqrt{2})((\sqrt{2})^2 - 2 ) = 0 \qquad (-\sqrt{2})((-\sqrt{2})^2 - 2 ) = 0\\
(0)(0-2) &= 0 \qquad\qquad (\sqrt{2})(2 - 2) = 0 \qquad\qquad\quad (-\sqrt{2})(2 - 2 ) = 0\\
(0)(-2) &= 0 \qquad\qquad\quad\ \ \  (\sqrt{2})(0) = 0 \qquad\qquad\qquad\ \ \  (-\sqrt{2})(0) = 0\\
0 &= 0 \qquad\qquad\qquad\qquad\ \ \   0 = 0 \qquad\qquad\qquad\qquad\qquad\ \ 0 = 0\\
\end{align}
$$
$$
\color{green}{True!}
$$
Everything works out.
Tips hat, walks away.
A: There is a third real answer.
You probably know how to factorise $a^2-b^2$ to $(a-b)(a+b)$
So with some imagination you can say $x(x-\sqrt2)(x+\sqrt2)=0$ implying that one of the following is true: 


*

*$x=0$ 

*$x-\sqrt2=0$ 

*$x+\sqrt2=0$. 

