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

The answers are $x=\sqrt{2}, 0$ how do I get there?

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  • $\begingroup$ Hint: "If at least one factor of a product is zero, then the product itself is equal to zero. The first factor is zero, if $x_1=0$. The second factor is zero, if $x^2-2=0$ Thus $x^2=2$. Taking the second root: $x_{2,3}=\pm \sqrt{2}$ Therefore you have three solutions. $\endgroup$ Commented Aug 29, 2015 at 9:05
  • $\begingroup$ There are various ways to define what real numbers are (e.g. Peano Axioms; ZFC set theory; that they are a complete ordered field), but they all give a result that if a.b = 0 then either a=0 or b=0. Take it from there. $\endgroup$ Commented Aug 29, 2015 at 15:28

7 Answers 7

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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 $x= 0 , \pm \sqrt{2}$

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    $\begingroup$ Like Henry’s, but better exaplained :)) $(+1)$ $\endgroup$
    – Mr Pie
    Commented Dec 15, 2017 at 6:44
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$1.$ 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 $x=-\sqrt 2$.

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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: $x = 0,-\sqrt{2}, \sqrt{2}$.

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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)

$\therefore x=0\space \mathrm{or} \space x^2-2=0$

$\implies x=0,\pm\sqrt2$

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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.

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  • $\begingroup$ Beautiful answer.... especially the very last sentence XD. $(+1)$ $\endgroup$
    – Mr Pie
    Commented Dec 15, 2017 at 6:40
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$x(x^2 - 2)=0$

implies $x=0$ or $x^2 - 2=0$

So $x=0$ or $x^2 = 2$.

So $x=0$, $x= \pm \sqrt{2}$

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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:

  • $x=0$
  • $x-\sqrt2=0$
  • $x+\sqrt2=0$.
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  • $\begingroup$ Creative :) $(+1)$ $\endgroup$
    – Mr Pie
    Commented Dec 15, 2017 at 6:42

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