Show that $x^3+3x+1=0$ has exactly one real solution.
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1Can you show some of your progress? Do you have shown the existence of one solution? – gerw Oct 13 '16 at 07:57
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1Wat happens when you take the limits $x\rightarrow \pm \infty$? Is the function continuous? – Mathematician 42 Oct 13 '16 at 07:59
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No i couldn't get it right if u know plz help me out – Manu Oct 13 '16 at 07:59
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2Did you cover Rolle's theorem in your lecture? – Dietrich Burde Oct 13 '16 at 08:02
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Use my comment to show the existence of a root and use the answer below to show that there is only one root. – Mathematician 42 Oct 13 '16 at 08:04
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Possible duplicate of Prove using Rolle's Theorem that an equation has exactly one real solution. – Dietrich Burde Oct 13 '16 at 08:07
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Also look into Descarte's Rule of Signs. – Sean Lake Oct 13 '16 at 08:43
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$f(x)=x^3+3x+1$ cannot have a positive real root and has atmost one negative real root (by Descarte's rule of signs). Since complex occur in pair, so one root has to be real. – Nitin Uniyal Oct 13 '16 at 14:22
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If the coefficients are all positive real values then it must have only one real root for any cubic equation. – Pentapolis Oct 13 '16 at 21:09
4 Answers
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Hint: use Calculus. Note that derivative always takes positive value.
P Vanchinathan
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I downvoted this. In my opinion you missed explaining key points. You should explain the significance of obtaining a positive derivative. It indicates that the slope of the curve is always increasing, and that it can cut the x-axis at only one point, hence it can have only one real root. – Akshar Gandhi Oct 13 '16 at 08:48
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@ Aksshar Gandhi. Quite often some answers posted here are hints (and I had clearly marked it as hint). – P Vanchinathan Oct 13 '16 at 09:01
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The discriminant is
$$\Delta = - 4 \cdot 3^3 - 27 = - 5 \cdot 27 < 0$$
From the negativity of $\Delta$, we conclude that the cubic polynomial has only one real root.
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A general method would be to check value of polynomial at Maxima and minima (Wiggles). If both values have same sign then only one real root. If no Wiggles then too single root.
jnyan
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Let $f$ defined by $f(x)=x^3+3x+1$. we know there is at least one real root.
assume there are $2$ real roots $a$ and $b$.
the polynomial function $f$ is continuous at $[a,b]$ and has a derivative in $(a,b)$.
Role's Theorem allows us to say there exists $c\in(a,b)$ such that
$f'(c)=3c^2+3=0$ which is not possible.
so there is only one real root.
hamam_Abdallah
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