# Determine the values of $k$ for which the equation $x^3-12x+k = 0$ will have three different roots.

Determine the values of $k$ for which the equation $x^3-12x+k = 0$ will have three different roots.

I can do this with calculus, I was just wondering what could be the pure algebraic approach to solve this?

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The three roots are distinct whenever the discriminant with respect to $x$, which in this case is $6912-27 k^2 = -27 (k-16) (k+16)$, is nonzero.

Or were you asking for three distinct real roots? The values of $k$ where it switches between having three real roots and having one real root will be zeros of the discriminant.

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There are formulas, which solve cubic equations: see for example the wikipedia article about cubic functions: here

With this explicit description of the zeros of your function, you can approach that question algebraically.

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When $k=0$, $x^3-12x=0\Leftrightarrow x=0\text{ or }x=\pm2\sqrt3$, so there are three real solutions in that case. Graphically, that means the curve $y=x^3-12x+k$ is "curvy enough" to have three distinct real zeros. The symmetry of the solutions in the $k=0$ case suggests that there are two values of $k$ at which we will go from having 3 real solutions, to 2 real solutions, to 1 real solution, and they are opposites. The boundary case, when we have 2 real solutions with one having multiplicity 2, should factor in the form $(x-a)(x-b)^2=x^3-(a+2b)x^2+(2ab+b^2)x-ab^2$, so we need $a+2b=0$ and $2ab+b^2=-12$. Solving that system gives $(a,b)=(4,-2)$ or $(a,b)=(-4,2)$, which give $k=-ab^2=\pm16$.

So, when $-16<k<16$, there are three distinct real solutions; when $k=\pm16$, there are two distinct real solutions; and when $k<-16$ or $k>16$, there is one real solution (and a conjugate pair of nonreal complex solutions).

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Suppose that the equation $f(x)=x^3-12x+k=0$ has a double root $y$. Then it easily follows that $f'(y)=0$, which is $3y^2-12=0$, i.e. $y=\pm 2$. So the only candidates for a double root are $\pm 2$.

If $y=2$ is a root of $f$ then $8-24+k=0$ and $k=16$.

If $y=-2$ is a root of $f$ then $-8+24+k=0$ and $k=-16$.

So when $k \neq \pm 16$ the roots of $f$ are distinct.

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We interpret "algebraic" somewhat loosely, in order to describe some material that was once a part of school algebra, but no longer is.

Look somewhat more generally at $x^3-ax+k=0$. Recall the identity $$\cos 3\theta=4\cos^3 \theta-3\cos\theta.$$ We want to make our cubic look like $4t^3-3t=c$.

Do this by making the change of variable $x=\lambda t$. Then our equation becomes $$\lambda^3 t^3-a\lambda t +k=0.$$ We want to have $\dfrac{\lambda^3}{a\lambda}=\dfrac{4}{3}$. Solve for $\lambda$. We get $\lambda=\sqrt{4a/3}$.

For simplicity we now go back to our specific numbers, though the argument can be made quite general. With $a=12$, we can take $\lambda=4$.

Substitute in our equation. After a bit of manipulation, it becomes $$4t^3-3t=-\frac{k}{16}.\qquad\qquad(\ast)$$ By comparison with the identity for $\cos 3\theta$, $(\ast)$ has three (not necessarily distinct) solutions iff $-k/16$ is the cosine of something (which we call $3\theta$). So the necessary and sufficient condition for $3$ not necessarily distinct real roots is $|-k/16|\le 1$. Let $3\theta=\arccos(-k/16)$. Then the solutions of $(\ast)$ are $\cos(\theta)$, $\cos(\theta+2\pi/3)$, and $\cos(\theta+4\pi/3)$. For some special values of $\theta$, two or more roots coincide.

More general cubics can be handled in the same way. If for example we are interested in $x^3+px^2+qx+r=0$, first make the substitution $x=y-p/3$. The resulting polynomial has no $y^2$ term.

The substitution trick $x=y-p/3$ goes back to Cardano. The trigonometric solution of the cubic in the case of all real roots is due to Vieta.

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