MHB Calculus involving cubic discriminants

[Dethrone]

My question is on this site:
https://ca.answers.yahoo.com/question/index?qid=20070217181026AAe29O6

There are two methods to do it, and I do not understand the first one in which the person uses cubic discriminants.

A cubic function is $ax^3+bx^2+cx+d=0$, and the function we are trying to find the cubic discriminants of is $4x^3-4x-(1+m)=0$. Therefore, $a=4$, $b=0$, $c=-4$, and $d=-(1+m)$. The poster on yahoo answers switches the roles of $b$ and $c$; how come he gets the correct answer whereas I didn't?

The cubic discriminant is given by:
$$\Delta=b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd$$

 

[MarkFL]

Let's take a look at the tangent lines to the given curve using our Desmos API:

[desmos="-3,3,-3,3"]y=x^4-2x^2-x;y=\left(4a^3-4a-1\right)(x-a)+a^4-2a^2-a;a=1[/desmos]

From this it would appear that for $a=\pm1$, we have the same tangent line.

Now, let's look at this analytically...the curve is:

$$y=x^4-2x^2-x$$

The tangent line is:

$$y=\left(4a^3-4a-1\right)(x-a)+a^4-2a^2-a$$

or in slope-intercept form:

$$y=\left(4a^3-4a-1\right)x-3a^4+2a^2$$

Let's let the two $x$-values of the tangent points be $a$ and $b$, where $a\ne b$. Since the slope of the two lines must be the same, we find:

$$4a^3-4a-1=4b^3-4b-1$$

This reduces to:

$$a^2+ab+b^2-1=0$$

Also, since the two intercepts must also be the same, we require:

$$-3a^4+2a^2=-3b^4+2b^2$$

And this reduces to:

$$(a+b)\left(3\left(a^2+b^2\right)-2\right)=0$$

Only the root $b=-a$ leads to distinct values, and so we find:

$$a^2+a(-a)+(-a)^2-1=0$$

$$a=\pm1$$

And thus the line:

$$y=-x-1$$

is the online line tangent to the given quartic at two distinct points. And the two points are:

$$(-1,0),\,(1,-2)$$

I know I didn't answer your actual question, but I thought I would share how I would work the problem.

 

[Dethrone]

Hi Mark (Poolparty),

How does $\displaystyle -3a^4+2a^2=-3b^4+2b^2$ simplify to $\displaystyle (a+b)\left(3\left(a^2+b^2\right)-2\right)=0$

 

[MarkFL]

Hi Mark (Poolparty),

How does $\displaystyle -3a^4+2a^2=-3b^4+2b^2$ simplify to $\displaystyle (a+b)\left(3\left(a^2+b^2\right)-2\right)=0$

Let's arrange it as:

$$3a^4-3b^4-2a^2+2b^2=0$$

Factor:

$$3\left(a^4-b^2\right)-2\left(a^2-b^2\right)=0$$

$$3\left(a^2+b^2\right)\left(a^2-b^2\right)-2\left(a^2-b^2\right)=0$$

$$\left(a^2-b^2\right)\left(3\left(a^2+b^2\right)-2\right)=0$$

$$(a+b)(a-b)\left(3\left(a^2+b^2\right)-2\right)=0$$

Since $a\ne b$, we may divide through by $a-b$:

$$(a+b)\left(3\left(a^2+b^2\right)-2\right)=0$$