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Recently I took an exam where it appeared the following equation:

$$x^3+x^2-1=0$$

On the exam we only needed to aprox. the root to the first decimal. However Wolframalpha says that there is a "computable" root.

I tried to work it out by trigonometric substitutions and the following:

$$(x^3+x^2-1)(x^3-x^2+1)=x^6-x^4+2x^2-1$$ Now:

$$x^2\to x$$

We get $$x^3-x^2+2x-1=0$$

From where I get nothing.

Is there anyway to work out manually the roots? And whether a third degree polynomial equation which you know that has a "computable" root can be manually worked out?

Thanks in advance.

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    You can use the Cardano's formula – 2017-01-20
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    Title isn't same with problem. – 2017-01-20
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    @arberavdullahu Look the computations... – 2017-01-20
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    Hm, I'm honestly not familiar with this method involving multiplying by another cubic. Do you happen to know it's name? – 2017-01-20
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    FYI, all third degree polynomials can be solved (or worked out), though you may run into casus irreducibilius, unless you pursue the solution involving trig functions like in my answer. – 2017-01-20

2 Answers 2

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Consider $f(x)=x^3+x^2-1$; then $f'(x)=3x^2+2x$, which vanishes at $-2/3$ and $0$. Therefore $f$ has a local maximum at $-2/3$ and a local minimum at $0$. Since $$ f(-2/3)=-\frac{8}{27}+\frac{4}{9}-1=\frac{-8+12-27}{27}=-\frac{23}{27} $$ we see that the equation has exactly one real solution. Since $f(0)=-1$ and $f(1)=1>0$, we know that this root is in the interval $(0,1)$.

For approximating the root we can use Newton’s method: $$ x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)} $$ starting at $x_0=1$. Then $$ x_1=1-\frac{1}{5}=\frac{4}{5}=0.8 $$ and $$ x_1=\frac{4}{5}- \frac{\dfrac{4^3}{5^3}+\dfrac{4^2}{5^2}-1} {3\dfrac{4^2}{5^2}+2\dfrac{4}{5}}= \frac{4}{5}-\frac{1}{5}\frac{64+80-125}{48+40}= \frac{4}{5}-\frac{19}{5\cdot 88}=\frac{333}{440}\approx0.7568 $$ Since the method provides alternatively an approximation by excess and one by defect, we can conclude the first decimal digit is $7$.

To check again, $$ f(0.7)=-0.167, \qquad f(0.8)=0.152 $$ so the first decimal digit is $7$.

Actually Cardan's formulas give you “explicitly” the root. Set $y=1/x$; then the equation is $y^3-y-1=0$; by Cardan's formula the only real root is $$ y=\sqrt[3]{\frac{1}{2}+\sqrt{\frac{1}{4}-\frac{1}{27}}} +\sqrt[3]{\frac{1}{2}-\sqrt{\frac{1}{4}-\frac{1}{27}}} \approx 1.32471795724474602596 $$ and $$ x=\frac{1}{y}\approx0.75487766624669276005 $$ According to bc, with this value for $x$ we get $$ f(x)=-0.00000000000000000001 $$ so it seems a pretty good approximation.

Your transformation can work as well, but it's more complex. Consider $$ g(x)=x^3-x^2+2x-1 $$ Then $g'(x)=3x^2-2x+2$ has negative discriminant, so the function $g$ is increasing. Its zero is between $0$ and $1$, and you can find it with a similar method. Then you need to find its positive square root.

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    :D And there is closed form for ya. – 2017-01-20
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    May I suggest Newton's method? I would've written an answer, but it would just be rewriting half of your answer. The reason I'm suggesting it because $x_{n+1} = x_n - \frac{x_n^3+x_n^2-1}{3x_n^2+2x_n}$ will give $\approx 0.7568$ in second iteration starting with $x_0 = 1$ (and you get $x_1 = 0.8$, so it is very doable by hands). – 2017-01-20
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    @Ennar I'd say go for a full numerical method answer, since this answer isn't really about numerical methods. – 2017-01-20
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    @Ennar You can surely suggest a better method! However, the iterative method starting with $x_0=1$ gives $0.707\dots$ at the first step, so the check is whether $f(0.7)<0$ and $f(0.8)>0$. – 2017-01-20
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    egreg, I wouldn't call it better, but I'd say easier to do by hand calculations than calculating a radical. – 2017-01-20
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    @Ennar I implemented it; the computations are indeed not overwhelming. – 2017-01-20
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If we start with your first line:

$$x^3+x^2-1=0$$

and let $x=\frac23y-\frac13$ and multiply both sides by $\frac{27}2$ so that we end up with

$$4y^3-3y-\frac{25}2=0$$

Recall that $4y^3-3y=\cosh(3\operatorname{arccosh}(y))$ so that we have

$$\cosh(3\operatorname{arccosh}(y))=\frac{25}2$$

and solving for $y$,

$$y=\cosh\left(\frac13\operatorname{arccosh}\left(\frac{25}2\right)\right)$$

$$x=\frac23\cosh\left(\frac13\operatorname{arccosh}\left(\frac{25}2\right)\right)-\frac13\approx0.754877666$$

and this is the only real root.

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    @Dr.MV See here: http://chat.stackexchange.com/rooms/20352/conversation/simple-arts-new-name – 2017-01-20
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    @Dr.MV He gave me a name I liked better is all :D – 2017-01-20
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    @Dr.MV See here: http://chat.stackexchange.com/rooms/51337/conversation/this-is-my-name-for-ya – 2017-01-21
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    @Dr.MV Lmao, I know that one. :D – 2017-01-21
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    You've heard that your whole life - and that gives you insurance. – 2017-01-21