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Let $ax^4 +bx^3 +cx^3 +dx + e = 0$ with $a,b,c,d,e\in\mathbb R$. I would like to know, how can I determine the condition for the polynomial to have exactly three distinct real solutions. one has to be a double root, or is there any other possibility.

I need help please.

Thanks

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  • $\begingroup$ Are there any restrictions? In particular, are $a,b,c,d,e$ real numbers? $\endgroup$
    – Calvin Lin
    Jun 11, 2013 at 14:28
  • $\begingroup$ they are real no.s $\endgroup$
    – Shobhit
    Jun 11, 2013 at 14:29
  • $\begingroup$ That is an extremely important assumption. Please edit it into your question. $\endgroup$
    – Calvin Lin
    Jun 11, 2013 at 14:33
  • $\begingroup$ If with "3 distinct roots" you mean the roots to be 3 distinct numbers a, b, c and one of them a double root, then it's possible. $\endgroup$
    – ypercubeᵀᴹ
    Jun 11, 2013 at 14:36

4 Answers 4

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Depending what you mean, if the equation has a double root eg $x^2(x+1)(x-1)=0$ it can have three distinct real roots, but we've counted one twice.

So there are two conditions:

(i) One double root (ii) The other two roots are real and distinct

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    $\begingroup$ You can computationally figure this out by computing the GCD of $f(x)$ and $f'(x)$. If there is a single double root, $r$, then the gcd should be $x-r$. Then factor out $(x-r)^2$ and determine if the resulting quadratic has real roots. $\endgroup$
    – Thomas Andrews
    Jun 11, 2013 at 14:40
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Hint: The complex conjugate root theorem (for polynomials with real coefficients) states that complex roots occur in conjugate pairs.

Hint: If we have 3 distinct real roots to a degree 4 equations, how many complex roots do we have?

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Usually roots are counted according to multiplicity of roots. So, in my humble opinion, it is assumed that first coefficient is zero: $a=0$. Thus we get cubic equation and there are conditions on discriminant for cubic equation such that it has 3 roots. See, for instance, http://en.wikipedia.org/wiki/Cubic_function

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Complex roots(in a polynomial with real coefficients) occur in pairs so the no. of real roots can be either $2$ or $4$

If there are 3 distinct real roots then the no. of real root(not necessarily distinct) is 4 hence there are no complex root.

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  • $\begingroup$ How about $x(x-1)(x-2)^2$? $\endgroup$
    – Martin Argerami
    Jun 11, 2013 at 14:34
  • $\begingroup$ There is no complex root of this equation.... $\endgroup$
    – Abhra Abir Kundu
    Jun 11, 2013 at 14:35
  • $\begingroup$ So? It's a degree $4$ polynomial with three distinct real roots. $\endgroup$
    – Martin Argerami
    Jun 11, 2013 at 14:35
  • $\begingroup$ one of the root has multiplicity 2 so it makes the equation to have 4 real roots and 3 $\huge{distinct}$ real roots $\endgroup$
    – Abhra Abir Kundu
    Jun 11, 2013 at 14:37
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    $\begingroup$ Which is what the OP asked: "polynomial to have exactly three distinct real solutions" $\endgroup$
    – ypercubeᵀᴹ
    Jun 11, 2013 at 14:38

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