Equation with a parameter

Question

Find m in order to have one real root in this equation: $$ 4x^3-12mx^2+m=0 $$ I tried Vieta's formula. I was thinking that: some equation $$ ax^3+bx^2+cx+d=0 $$

$$ s_1=x_1+x_2+x_3=\frac{-b}{a} $$ $$ s_2=x_1x_2+x_2x_3+x_1x_3=\frac{c}{a} $$ $$ s_3=x_1x_2x_3=\frac{-d}{a} $$ square the first relation $$x_1^2+x_2^2+x_3^2=s_1^2-2s_2$$ if$$ s_1^2-2s_2<0 \implies $$ there exist a pair of complex roots (i assume that the complex roots are in pairs $$z_1,z_2;\;\; z_1=\overline z_2 $$)=> there is only one real root. But when I plug in with $$ m $$ $$s_2=0$$ $$ 9m^2 < 0 \implies m \in \emptyset $$ And there is no answer like that.

Thank you!

Answer 1

Write $$m={4x^3\over 12x^2-1}$$

So we are interested for which real $m$ the line $y=m$ cuts graph of $$f(x)={4x^3\over 12x^2-1}$$

Let $a=\sqrt{1\over 12}$

Since $$\lim_{x\to -a_+} = \infty\;\;\;\;\;\;\;{\rm and}\;\;\;\;\;\;\;\lim_{x\to a_-} = -\infty $$ and $f$ is continous on $(-a,a)$ we have $Im(f)=\mathbb{R}$, so $m \in \mathbb{R}$.

Answer 2

HINT: $$m=\frac{4x^3}{12x^2-1}$$ and define $$f(x)=\frac{4x^3}{12x^2-1}$$