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A cubic equation $ax^3+bx^2+cx+d=0 \space$ where, $a\neq 0$ always has one real root.

Is there any direct condition for determining the nature i.e. sign of one real root for sure?

Is it possible by simply observing only the signs of coefficients without making any calculation?

Harish Chandra Rajpoot
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2 Answers2

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Let $$f(x)=ax^3+bx^2+cx+d$$ $$f'(x)=3ax^2+2bx+c$$ The cubic equation will have exactly one real root if $\mathrm{Discriminant \space of \space f'(x)}<0$. (Sufficient but not necessary condition).

In that case the cubic equation will be strictly increasing or decreasing according as $a$ is positive or negative.

In that case, to check if your root is positive or negative, simply see the sign of $f(0)=d$.

For the case when $a>0$,

If $d>0$, the curve will have intersected the x-axis before $0$ and the root is negative and vice versa for the other case.

Arpan
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    It also depends on $a>0$ or $a<0$, but it's not restrictive to assume $a>0$ or even $a=1$. More important is that the condition that $f'(x)$ has negative discriminant is only sufficient for the uniqueness of the real root. – egreg Apr 23 '15 at 14:10
  • @egreg You're right, I've included it now. – Arpan Apr 23 '15 at 14:12
  • Note that $f(x)=x^3-x+10$ has only one real root, but the discriminant of $f'(x)$ is positive. – egreg Apr 23 '15 at 14:14
  • As the OP asks for one, not the, real root, you are not restricted to the case you consider. The Sign of $a$ and $d$ tells you the sign of one real root. – martini Apr 23 '15 at 14:15
  • I assumed the OP was asking for the case when it has only one real root. – Arpan Apr 23 '15 at 14:16
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Suppose you know there is exactly one real root, and suppose without loss of generality that $a>0$ (it isn't zero, so if it's negative, multiply both sides by $-1$ and work with that equation instead).

Call the expression on the left side $f(x)$. Note that $f$ changes sign exactly once (at the root), so it must be positive on one side of the root and negative on the other. And since $a>0$, the function is negative for large negative $x$ and positive for large positive $x$ (that is, the function runs roughly uphill to the right, "in the large", if we ignore the possible wiggle it may have somewhere in between). Note also that $f(0) = d$.

If $d=0$, the root is zero and you are done. Otherwise either $d>0$, so that the root must be to the left of $0$ (i.e., the root is negative), or $d<0$, so that the root is to the right of $0$ (i.e., the root is positive).

The gist of the idea here is that there is a point on the graph of $f$ in the 3rd quadrant, and there is a point on the graph in the first quadrant. The graph connects these points, and must cross the $y$-axis somewhere. If it crosses the positive $y$-axis, then it must also cross the negative $x$-axis. If it crosses the negative $y$-axis, then it must also cross the positive $x$-axis. Think of trying to draw the graph of any continuous function that passes through the three points and you will be convinced that you must cross the $x$-axis in the manner described.

MPW
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