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The problem is as follows

If $x^2+x-1=0$, compute all possible values of $\frac{x^2}{x^4-1}$

This was a no-calculator 10 min for 2 problem format contest.

I started by using quadratic formula, but the answers I got were too ugly to be plugged into the equation before time ran out. I tried several algebraic manipulations, but after losing time on the other problem, i found no quick answer to this. I presume there is one step i'm missing to make that leap. Can this be done easily?

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  • $\begingroup$ $x^4-1$ factors out to $(x^2+1)(x+1)(x-1)$. $\endgroup$ – Jason Chen May 18 '14 at 5:02
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$$x^2=1-x$$

$$\implies\frac{x^2}{x^4-1}=\frac{1-x}{(1-x)^2-1}=\frac{1-x}{x^2-2x}=\frac{1-x}{1-x-2x}$$

$$=\frac{1-x}{1-3x}\ \ \ \ (1)$$

$$=\frac13\left(1+\frac2{1-3x}\right)\ \ \ \ (2)$$

Now put the two values of $x$, one by one in either $(1)$ or in $(2)$ which ever you like

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  • $\begingroup$ Regardless of how practical this may be in the short time-frame I had, I respect its compacting into one equation. $\endgroup$ – Asimov May 18 '14 at 5:11
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Use the quadratic formula and solve for the zeros.

Verify that they are $x = \frac{1}{2}(-1-\sqrt{5}),\frac{1}{2}(\sqrt{5}-1)$.

Now just evaluate the later for these values of $x$.

$\textbf{Extra}$: You can also complete the square and you will get: $(x+\frac{1}{2})^2 -\frac{5}{4}=0$, then solve for $x$.

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  • $\begingroup$ Taking that to the fourth power was what was giving me difficulty, I had gotten that far, and was wondering if there was a different route. $\endgroup$ – Asimov May 18 '14 at 5:03
  • $\begingroup$ Yes, complete the square and you will get: $(x+\frac{1}{2})^2 -\frac{5}{4}=0$, then solve for $x$. $\endgroup$ – Mr.Fry May 18 '14 at 5:04
  • $\begingroup$ I meant after i got those values, if there was an easier way to find $\frac{x^2}{x^4-1}$ without having to find x. An algebraic way around the ugly part. $\endgroup$ – Asimov May 18 '14 at 5:06
  • $\begingroup$ Oh yes: As @Jason Chen says, $(x^4-1)=(x^2+1)(x+1)(x-1)$. I hope this answered your question :). $\endgroup$ – Mr.Fry May 18 '14 at 5:08
  • $\begingroup$ Ok, I know how to do that, I was just wondering if there was a concise shortcut that removed that part entirely. I guess there may not be in this case. $\endgroup$ – Asimov May 18 '14 at 5:08
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Let $a, b$ be the roots of $x^2+x-1=0$. Then $a+b = -1, ab = -1$. For simplicity, consider now the reciprocals of the values we wish to find, i.e. $\dfrac{a^4-1}{a^2} = k_1$ and $\dfrac{b^4-1}{b^2} = k_2$, then

$$k_1 + k_2 = a^2+b^2-\left(\frac1{a^2}+\frac1{b^2} \right), \quad k_1 k_2 = \frac{(a^4-1)(b^4-1)}{(ab)^2}$$

Now $a^2+b^2 = (a+b)^2-2ab = 3$, so $k_1 + k_2 = 3-3=0$.
Also $(a^4-1)(b^4-1) = 1+(ab)^4-(a^4+b^4) = 1+1-(3^2-2) = -5$, so $k_1 k_2 = -5$.

So $k_1, k_2$ are roots of $y^2-5=0 \implies \{k_1, k_2\} = \{\sqrt5, -\sqrt5\}$. Take reciprocals now.

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If $x^2+x-1=0$, then $x^2=1 – x$

Squaring both sides, we get $x^4=1 – 2x + x^2$

∴ $x^4 – 1 = x^2 – 2x$

$= (1 – x) – 2x$

$= 1 – 3x$

Thus, the expression $=$ ${1 – x} \over {1 – 3x}$

By long division, it is equivalent to ${1} \over {3}$+ ${2}\over {3(1 – 3x)}$

Edit (1) - Please ignore the following:-

As pointed out, $x^4 – 1$ can also be factorized as $(x^2+1)(x+1)(x-1)$.

Thus, the expression can take any value of x except 1, -1, or 1/3 (if real values of x are assumed).

Edit (2) - Please add the following:-

As pointed out, $x = \frac{1}{2}(-1 \pm \sqrt{5})$

Just put $x = \frac{1}{2}(-1 + \sqrt{5})$ in the simplified expression and do a little rationalization first. For the $x = \frac{1}{2}(-1 - \sqrt{5})$, the result is very similar.

Using the simplified expression, 10 minutes should be sufficient.

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  • $\begingroup$ I didn't downvote, but $x$ can only take two values, so this expression can only take two values. $\endgroup$ – Empy2 May 18 '14 at 5:37
  • $\begingroup$ Solution has been edited. Thanks for pointing that out. $\endgroup$ – Mick May 18 '14 at 8:53
  • $\begingroup$ You cannot square both sides before you have proved that $1-x$ takes non-negative values only. $\endgroup$ – user26486 May 18 '14 at 15:29
  • $\begingroup$ I think squaring both sides is allowable (but after that its reverse process should not be conducted because extraneous root(s) will be generated). $\endgroup$ – Mick May 19 '14 at 4:17
  • $\begingroup$ I'm not sure what you mean. Consider the equation $f(a)=f(b)$. Then according to you $f^2(a)=f^2(b)\implies f(a)=f(b)$, but consider, e.g., $3=-3$ and you'll see where you're wrong. And please add @mathh to your comment addressing me since I won't be able to see it otherwise; you are lucky I've accidentally stumbled across this question again. $\endgroup$ – user26486 May 24 '14 at 2:49

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