MHB Can you solve these quadruple equations with real numbers?

[MarkFL]

Hello, MHB Community! (Wave)

anemone has asked for me to fill in for her while she's away on holiday. :)

Here is this week's POTW:

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Find all quadruples $(a,b,c,d)$ of real numbers that simultaneously satisfy the following equations:$$\left\{\begin{array}{rcl}a^3+c^3 & = & 2 \\ a^2b+c^2d & = & 0 \\ b^3+d^3 & = & 1 \\ ab^2+cd^2 & = & -6 \end{array}\right.$$

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[MarkFL]

Hello, MHB Community! (Wave)

It has been brought to my attention that the problem posted this week has been used in the past, found here:

https://mathhelpboards.com/potw-secondary-school-high-school-students-35/problem-week-277-aug-28th-2017-a-22196.html

I apologize for that...here is the solution with which I was provided:

Spoiler

Consider the polynomial $P(x)=(ax+b)^3+(cx+d)^3=\left(a^3+b^3\right)x^3+3\left(a^2b+c^2d\right)x^2+3\left(ab^2+cd^2\right)x+b^3+d^3$. By the conditions of the problem, $P(x)=2x^3-18x+1$. Clearly $P(0)>0,\,P(1)<0$ and $P(3)>0$. Thus $P$ has three distinct zeroes. But $P(x)=0$ implies $ax+b=-(cx+d)$ or $(a+c)x+b+d=0$. This equation has only one solution, unless $a=-c$ and $b=-d$. But since the conditions of the problem do not allow this, we infer that the system of equations in the problem has no solution.

So, I am going to post another problem, from my old physics homework.

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A uniform rod of mass $M$ and length $d$ rotates in a horizontal plane about a fixed, vertical, frictionless pin through its center. Two small beads, each of mass $m$, are mounted on the rod such that they are able to slide without friction along its length. Initially the beads are held by catches st positions $x$ (where $x<d/2$) on each side of the center, at which time the system rotates with an angular speed $\omega$.

Suddenly the catches are released and the small beads slide outward along the rod. Find:

• (a) the angular speed of the system at the instant the beads reach the ends of the rod
• (b) the angular speed of the rod after the beads fly off the ends

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[MarkFL]

No one answered this weeks POTW...my solution follows:

Spoiler

(a) Apply conservation of angular momentum (subscripts of $R$ pertain to the rod, while those of $B$ pertain to the beads, and $\omega_i=\omega$):

Recall angular momentum $L$ is the product of the moment of interia $I$ and the angular speed $\omega$.

$$I_R\omega_i+2I_{B_i}\omega_i=I_R\omega_f+2I_{B_f}\omega_f$$

Solve for $\omega_f$:

$$\omega_f=\omega_i\frac{I_R+2I_{B_i}}{I_R+2I_{B_f}}$$

Now the moment of inertia for a point mass (which we can use for the small beads where no radius is given) of mass $m$ and distance from the axis of rotation $r$ is:

$$I=mr^2$$

And the moment of rotation for a rod of mass $m$ and length $L$ is:

$$I=\frac{1}{12}mL^2$$

And so, there results:

$$\omega_f=\omega_i\frac{\frac{1}{12}Md^2+2mx^2}{\frac{1}{12}Md^2+2m\left(\frac{d}{2}\right)^2}=\omega_i\frac{Md^2+24mx^2}{Md^2+6md^2}$$​

(b) We can equate the momentum of the system when the beads are at the ends of the rod to when the beads have flown off the rod as follows:

$$d^2\omega_f\left(\frac{1}{12}M+2m\right)=I_R\omega_R+2mv_B$$

Now, observing that $$v_B=\frac{d}{2}\omega_f$$, and substituting for the moment of inertia for the rod, we have:

$$d^2\omega_f\left(\frac{1}{12}M+2m\right)=\frac{1}{12}Md^2\omega_R+md\omega_f$$

Solve for $\omega_R$:

$$\omega_R=\frac{\omega_f}{Md}\left(d\left(M+24m\right)-12m\right)$$​