Show that the roots of the equation $b^2(x^4+1)-14b(1+b^2)x(x^2+1)+(1+196b^2+b^4)x^2=0$ are all real and all have the same sign as $b$.

Question

If $\alpha, \beta$ are the roots of the equation $$x^2-ax+b=0\ (a,b\ real)$$ form the equation whose roots are $\frac{\alpha^3}{\beta},\frac{\beta^3}{\alpha}$, and deduce the equation whose roots are $\frac{\alpha}{\beta^3},\frac{\beta}{\alpha^3}$.

Show that the roots of the equation $$b^2(x^4+1)-14b(1+b^2)x(x^2+1)+(1+196b^2+b^4)x^2=0$$ are all real and all have the same sign as b. $$\\$$

My Work

Here, I have worked out the answer for roots $\frac{\alpha^3}{\beta},\frac{\beta^3}{\alpha}$ is $$bx^2-(a^4-4a^2b+2b^2)x+b^3=0$$

And the answer for roots $\frac{\alpha}{\beta^3},\frac{\beta}{\alpha^3}$ is $$b^3x^2-(a^4-4a^2b+2b^2)x+b=0$$

But, to prove the roots are real for $$b^2(x^4+1)-14b(1+b^2)x(x^2+1)+(1+196b^2+b^4)=0$$ I have rearranged the equation to become $$b^2x^4-(14b+14b^3)x^3+(1+196b^2+b^4)x^2-(14b+14b^3)x+b^2=0$$ And by letting $\alpha,\beta,\gamma,\delta$ be roots for the quartic equation. I come up with this: $$\alpha+\beta+\gamma+\delta=\frac{14b+14b^3}{b^2}=\frac{14}{b}+14b$$ $$\alpha\beta\gamma\delta = 1$$ But these aren't giving me any clues to continue the work.
Can anyone give me some clues for this?

Answer 1

We have $$b^2x^4-14(b^3+b)x^3+(b^4+196b^2+1)x^2-14(b^3+b)x+b^2=0$$ or, better $$(x^2-14bx+b^2) [(bx)^2-(14bx)+1) ]=0$$ or $$\left[\left(\dfrac xb\right)^2-14\left(\frac xb\right)+1\right][(bx)^2-14(bx)+1]=0$$

whose roots are $\dfrac xb=7\pm4\sqrt3$ and $bx=7\pm4\sqrt3$. Since $7\gt4\sqrt3$ the two roots are positive in both cases so the sign of the four roots $x$ are the same that the sign of $b$.

Answer 2

If $b = 0$, then the equation becomes $x^2 = 0$.

In the following, assume that $b \ne 0$.

Since $x = 0$ is not a root, we multiply both sides of the equation by $\frac{1}{b^2x^2}$ to get $$\left(x^2+\frac{1}{x^2}\right) -14\left(b + \frac{1}{b}\right)\left(x+\frac{1}{x}\right) + \left(b^2 + \frac{1}{b^2} + 196\right) = 0$$ or $$\left(\left(x+\frac{1}{x}\right)^2 - 2\right) -14\left(b + \frac{1}{b}\right)\left(x+\frac{1}{x}\right) + \left(\left(b + \frac{1}{b}\right)^2 + 194\right) = 0$$ or $$\left(y^2 - 2\right) -14cy + \left(c^2 + 194\right) = 0$$ or $$y^2 - 14cy + c^2 + 192 = 0 \tag{1}$$ where $y = x + \frac{1}{x}$ and $c = b + \frac{1}{b}$.

The discriminant of (1) is $192(c^2 - 4)\ge 0$ (easy). Thus, (1) has two real roots. According to Vieta's theorem, if $c > 0$, (1) has two positive real roots; and if $c < 0$, (1) has two negative real roots. Also, (1) can be written as $(c - 7y)^2 = 48(y^2 - 4)$. Thus, all roots of (1) satisfy $y^2 \ge 4$. Thus, $x + \frac{1}{x} = y$ always has two real roots. Thus, the original quartic equation has four real roots.

Note that $x$ and $y$ have the same sign, and $c$ and $b$ have the same sign. Thus, all four real roots have the same sign as $b$.