Find all values of $a$ for which there are two real solutions of $x^3-2ax^2+a^2x-3=0$

Question

Find all values of $a$ for which there are two real solutions of the equation.

$$x^3-2ax^2+a^2x-3=0$$

Ans = $1.5\sqrt[3]{6}$

I tried to research the function by dint of derivative, but it didn't give me anything.

Answer 1

Try to plot a graph of a cubic function. In your case it's $-\infty$ on the left, going concave, then going convex and then increasing to $+\infty$.

A cubic function has three (with multiplicity) roots. If two are real, then the third is real, too (hint: coefficient at $x^0$), therefore, two roots must coincide, but not all three.

Hence, our function must have the form $(x-b)(x-c)^2$, $b\ne c$. By looking at the derivatives of zeroth, first, and second order, you can build a system of equations on $a$, $b$, $c$ and obtain the expression for $a$.

Another approach, as suggested by @gnasher729, is to simply open the parantheses and match coefficients at different powers of $x$.

Answer 2

The correct way is to consider the equation as an equation for $a$ first. In this way you will get $$xa^2-2x^2a+x^3-3=0$$

Then $D = x^4-x(x^3-3)=3x$ and $a_1=\frac{x-\sqrt{3x}}{x}$ and $a_1=\frac{x+\sqrt{3x}}{x}$ then you can factor the equation as follows $x(a-\frac{x-\sqrt{3x}}{x})(a-\frac{x+\sqrt{3x}}{x})=0$. Now you have two equations

$xa-x+\sqrt{3x}=0$

$xa-x-\sqrt{3x}=0$

Answer 3

As complex roots come in conjugates, there is a repeated root. This means there exists a $x_0$ where $f(x_0)=f'(x_0)=0$

Solving $f'(x_0)=0$ gives a quadratic with roots $a,a/3$

As function is non zero at $a$, this implies $f(a/3)=0\implies a^3(1/27-2/9+1/3)-3=0$

This gives you the desired result.