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I want to solve a polynomial equation but I know that it can have exactly one root. Is there some method to solve these kinds of problems.

for example- $$A(1+x)^4 + B(1+x)^3 +C(1+x)^2 + D(1+x) +E=0$$

Here $A$ is always negative, and $B,C,D,E$ are always positive.

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  • $\begingroup$ "Exactly one root" where? This one will always have at least two real roots, one $< -1$ and one $> -1$. $\endgroup$ Commented Jul 10, 2016 at 20:45

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Let $ f(x)=A(1+x)^4 + B(1+x)^3 +C(1+x)^2 + D(1+x) +E=0$

Since $A$ is negative, $f(-\infty) = f(\infty)=-\infty$ and if this polynomial has just one root that should be a double root. Therefore you can take $gcd(f(x),f'(x))$.

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  • $\begingroup$ I did not understand, since x is unknown how can I take gcd of f(x) and f'(x) ? $\endgroup$
    – Noober
    Commented Jul 10, 2016 at 7:30
  • $\begingroup$ by application of extended Euclidean algorithm on f(x) and f'(x) $\endgroup$
    – xyz
    Commented Jul 10, 2016 at 7:35
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A quartic equation such as this can be solved using radicals, but it's rather complicated. If you have numerical values for $A,B,C,D,E$ you're probably better off using numerical methods.

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  • $\begingroup$ Hi, thanks for the answer. What do you mean by numerical methods? $\endgroup$
    – Noober
    Commented Jul 10, 2016 at 7:10
  • $\begingroup$ @Noober He means approximation techniques using calculus, such as Newton's method or a variety of others. $\endgroup$
    – JasonM
    Commented Jul 10, 2016 at 7:12
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Having exactly one root means exactly one thing: the following equation has the form $A(x - r)^4$. So you need to write this in expanded form:

$$ A(x - r)^4 = Ax^4 - 4Ax^3r + 6Ax^2r^2 - 4Axr^3 + Ar^4 $$

then we have:

\begin{align} \alpha(1+x)^4 + \beta(1 + x)^3 + \gamma(1 + x)2 + \delta(1 + x) + \epsilon =&\ \alpha x^4 + 4\alpha x^3 + 6\alpha x^2 + 4\alpha x + \alpha \\ &\ + \beta x^3 + 3\beta x^2 + 3\beta x + \beta \\ &\ + \gamma x^2 + 2\gamma x + \gamma \\ &\ + \delta x + \delta \\ &\ + \epsilon \\ =&\ \alpha x^4 + (4\alpha + \beta)x^3 + (6\alpha + 3\beta + \gamma)x^2 + (4\alpha + 3\beta + 2\gamma + \delta)x + (\alpha + \beta + \gamma + \delta + \epsilon) \end{align}

This creates five equation with 6 uknowns (although I have a feeling $A$ can be set to equal a multiple of any of the $\alpha, \beta, \gamma, \delta, \epsilon$).

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