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How do I solve $2x^4-3x^2+1=0$, for $x$?
Or any idea how to turn it into a quadratic equation?

beep-boop
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jackcall
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  • Do you notice that 1 and -1 are roots of this equation? Thus, you could divide it by $x^2-1$ to try to find the other 2 factors. – JB King Apr 03 '14 at 17:23

8 Answers8

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Equations of the form $ax^{2n}+bx^n+c=0$ are called biquadratics and can be seen as a quadratic by setting $t=x^n$ and then solving for $t$.

In your specific example we would set $t=x^2$ and then solve $2t^2-3t+1=0$ which gives $t=1\lor t=\frac{1}{2}$. Now recall that $t=x^2$ so we need to solve $x^2=1$ and $x^2=\frac{1}{2}$ which can be easily solved

Alessandro Codenotti
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Hint: Substitute $t=x^2{}{}{}{}{}{}{}{}{}$

Guy
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  • I like how people posted the same answer before you and still you got more upvotes. – evil999man Apr 03 '14 at 17:25
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    @Awesome With minimal effort. :D If you click "edit" you will see I had to use a standard hack to overcome the minimum 30 char restriction as well. – Guy Apr 03 '14 at 17:56
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Let $t=x^2$ so we have $$2t^2-3t+1=0\iff t_{1}=1,\;t_2=\frac12$$ hence $$\text{the set of solutions }=\left\{\pm1,\pm\frac1{\sqrt2}\right\}$$

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$$\begin{eqnarray} &&\ 2x^4\,\ -\ 3\,\ x^2\,\ +\ 1\, =\, 0\\ \smash[t]{\overset{\large \times\, 2}\iff} &&\ 4x^4-3(2x^2)\,+\,2\, =\, 0\\ \iff &&\ \ X^2 -\ 3\,\ X\ \ +\ \ 2\, =\, 0,\ \ X = 2x^2\\ \iff && \ (X\ -\ 2)\, (X\ -\: 1) =\, 0 \end{eqnarray}$$

Remarks $\ $ This method works generally to reduce to polynomials that are monic, i.e. have leading coefficient $= 1,\,$ see the AC-method.

Community
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Bill Dubuque
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by substituing x^2=y you can turn it into quadratic equation.

SA-255525
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Hint: Change the variable $x^2$ by a new variable $t$. So put $x^2=t$. You will get $$2t^2-3t+1=0.$$ Solve it in $t$ and return to your variable $x$.

x.y.z...
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Alternatively, we can see that the factors of each term in the polynomial $2x^4-3x^2+1$ add up to $0$. And the polynomial have only even coefficients, so $\pm1$ is a solution. Thus, we can divide this polynomial by $(x-1)(x+1)=x^2-1$ using Euclidean division to find: $2x^2-1$, and the remainder is of course $0$.

We could therefore write $2x^4-3x^2+1$ as $(x-1)(x+1)(2x^2-1)$ which can be factored even further by noting that $2x^2-1=(\sqrt{2}x)^2-1^2=(\sqrt{2}x-1)(\sqrt{2}x+1)$. So the additional roots are determined by solving $\sqrt{2}x\pm1=0$ which is equivalent to $x=-1/\sqrt{2}$ or $x=1/\sqrt{2}$.

We conclude that the set of solutions to the equation $2x^4-3x^2+1=0\,\,(\star)$ is: $$\text{Set of solutions to $(\star)$}=\left\{1,-1,-\frac1{\sqrt{2}},\frac1{\sqrt{2}}\right\}.$$

I hope this helps.
Best wishes, $\mathcal H$akim.

Hakim
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Put t=x^2 then solve the quadratic equation in t.

Gurjot Singh
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    This question is 2 months old, and most of the other 7 answers are essentially identical to yours. I think it would be a good idea for you to either improve this answer with additional insights, or delete it. Next time, please check to see if you are bringing something new to the discussion before bumping old questions. – epimorphic Jun 07 '14 at 02:12