Find all values of m.

anemone

MHB POTW Director
Staff member
Problem:
Find all values of $m$ for which the equation $16x^4-mx^3+(2m+17)x^2-mx+16=0$ has four distinct roots forming a geometric progression.

Attempt:
We're told that the four distinct roots of the equations above forming a geometric progression. So, I let the roots of the equation above as $a, ar, ar^2$, and $ar^3$ where $a$ and $r$ are the first term and common ratio of the geometric progression respectively.

These are the three equations that I determined by using Vieta's formulas and and I can't see where these equations could lead me to the values of m...

$\displaystyle a^2r^3=1$

$\displaystyle a(1+r)(1+r^2)=\frac{m}{16}$

$\displaystyle\left(\frac{(1+r)}{r}\right)^2(r^2-r+1)=\frac{2m-15}{16}$

The farthest that I could go in the effort of finding the values of m by blending all these equations together is the following:

$\displaystyle \left(\frac{(1+r)}{r}\right)^2\left[\frac{(r^2-r+1)(r^2+r+1)}{r^2}+1\right]=\frac{m^2}{16}$

and as you can see, this doesn't seem right because I couldn't eliminate either the variables $m$ or $r$ to solve for either of them and thus, the values of m remain unknown.

chisigma

Well-known member
Problem:
Find all values of $m$ for which the equation $16x^4-mx^3+(2m+17)x^2-mx+16=0$ has four distinct roots forming a geometric progression.

Attempt:
We're told that the four distinct roots of the equations above forming a geometric progression. So, I let the roots of the equation above as $a, ar, ar^2$, and $ar^3$ where $a$ and $r$ are the first term and common ratio of the geometric progression respectively.

These are the three equations that I determined by using Vieta's formulas and and I can't see where these equations could lead me to the values of m...

$\displaystyle a^2r^3=1$

$\displaystyle a(1+r)(1+r^2)=\frac{m}{16}$

$\displaystyle\left(\frac{(1+r)}{r}\right)^2(r^2-r+1)=\frac{2m-15}{16}$

The farthest that I could go in the effort of finding the values of m by blending all these equations together is the following:

$\displaystyle \left(\frac{(1+r)}{r}\right)^2\left[\frac{(r^2-r+1)(r^2+r+1)}{r^2}+1\right]=\frac{m^2}{16}$

and as you can see, this doesn't seem right because I couldn't eliminate either the variables $m$ or $r$ to solve for either of them and thus, the values of m remain unknown.

First we can write the equation in a different form...

$\displaystyle x^{4} - \frac{m}{16}\ x^{3} + \frac{2m+17}{16}\ x^{2} - \frac{m}{16}\ x + 1 =0$ (1)

... then we indicate the four roots as $x_{1}= a,\ x_{2}= a\ r,\ x_{3}= a\ r^{2},\ x_{4}= a\ r^{3}$ and write the (1) in the form...

$\displaystyle (x-x_{1})\ (x-x_{2})\ (x-x_{3})\ (x-x_{4})=0$ (2)

Now if we impose that the coefficients of the term in $x^{0}$, $x$ and $x^{3}$ in (1) and (2) are the same we obtain...

$\displaystyle a^{4}\ r^{6}=1$

$\displaystyle a^{3}\ (r^{3}+ r^{4} + r^{5} + r^{6})= \frac{m}{16}$

$\displaystyle a\ (1 + r + r^{2} + r^{3}) = \frac{m}{16}$ (3)

The (3) is a system of three equations in m, a and r which can be solved. If a solution of (3) satisfies also the condition on coefficient of the term $x^{2}$, then this solution is 'good' ... if not the problem doesn't have solutions...

Kind regards

$\chi$ $\sigma$

Last edited:

Opalg

MHB Oldtimer
Staff member
I have a feeling that Vieta's relations are not going to be the best way to tackle this problem (though I admit I don't have a better method). I started by writing the equation as $$16x^4 +17x^2+16= mx(1-x)^2. \qquad(*)$$ You should be able convince yourself that (*) will only have four real solutions if $m>0.$ In that case, since $x(1-x)^2$ is zero when $x=0$ or $1$, it seems clear that two of the solutions must lie in the interval $(0,1)$ and the other two in the interval $x>1$. But the maximum value of $x(1-x)^2$ in the interval $(0,1)$ is $\frac{16}{27}$, and the function $16x^4 +17x^2+16$ is greater than $16$ throughout that interval. So it is clear that $m$ will have to be quite large for (*) to have any solutions at all in that interval.

I found by trial and error, using a graphing calculator, that (*) has a solution of the required form when $m=170$. In fact, $$16x^4 +17x^2+16 - 170x(1-x)^2 = (8x-1)(2x-1)(x-2)(x-8),$$ with solutions $\frac18,\ \frac12,\ 2,\ 8$ in geometric progression. I think that must be the unique solution to the problem, but I do not see any way of deriving that result analytically.

anemone

MHB POTW Director
Staff member
I have a feeling that Vieta's relations are not going to be the best way to tackle this problem (though I admit I don't have a better method). I started by writing the equation as $$16x^4 +17x^2+16= mx(1-x)^2. \qquad(*)$$ You should be able convince yourself that (*) will only have four real solutions if $m>0.$ In that case, since $x(1-x)^2$ is zero when $x=0$ or $1$, it seems clear that two of the solutions must lie in the interval $(0,1)$ and the other two in the interval $x>1$. But the maximum value of $x(1-x)^2$ in the interval $(0,1)$ is $\frac{16}{27}$, and the function $16x^4 +17x^2+16$ is greater than $16$ throughout that interval. So it is clear that $m$ will have to be quite large for (*) to have any solutions at all in that interval.

I found by trial and error, using a graphing calculator, that (*) has a solution of the required form when $m=170$. In fact, $$16x^4 +17x^2+16 - 170x(1-x)^2 = (8x-1)(2x-1)(x-2)(x-8),$$ with solutions $\frac18,\ \frac12,\ 2,\ 8$ in geometric progression. I think that must be the unique solution to the problem, but I do not see any way of deriving that result analytically.
Many thanks, Opalg.

I want you to know how much I admire your intelligence in maths and how much I appreciate your kindness in spending time to help me with my problems and typing out your solutions in such a lucid and easy to read manner.

Thanks again.

Opalg

MHB Oldtimer
Staff member
Many thanks, Opalg.

I want you to know how much I admire your intelligence in maths and how much I appreciate your kindness in spending time to help me with my problems and typing out your solutions in such a lucid and easy to read manner.

Thanks again.
Thank you for those kind comments. In return, I want you to know how much I enjoy the problems that you post. They are always interesting, unusual, challenging and rewarding to solve. I wonder where you get them from?

anemone

MHB POTW Director
Staff member
Thank you for those kind comments. In return, I want you to know how much I enjoy the problems that you post. They are always interesting, unusual, challenging and rewarding to solve. I wonder where you get them from?
Hi Opalg, that's so nice of you to say so. My passion for maths has developed since I was 17 years old and solving all hard and difficult problems could let me feel somehow closer to my papa in my heart and as such, I'm always hunting for hard problems (that's compatible to my knowledge in maths) everywhere, some are from the collection of Olympiad maths problems, whereas other are from USA mathematical talent search's site, and etc.

I'll find one of the sites and forward the link to you in a PM soon.

anemone

MHB POTW Director
Staff member
I have just found another method online on how to solve this problem and I want to share it with everybody here...

We're asked to find all values of $m$ for which the equation $16x^4-mx^3+(2m+17)x^2-mx+16=0$ has four distinct roots forming a geometric progression.

Note that $$\displaystyle x=0$$ is not a root of the equation. Then we can write it as $16x^2-mx+(2m+17)-\frac{m}{x}+\frac{16}{x^2}=0$ or equivalently,

$16t^2-mt+(2m-15)=0$ where $$\displaystyle t=x+\frac{1}{x}$$ and hence $$\displaystyle |t|\ge 2$$.

From this it follows that the given equation has four distinct real roots if and only if the quadratic function $f(t)=16t^2-mt+(2m-15)$ has two distinct real roots
$$\displaystyle t_1,\;t_2$$ which are not in $$\displaystyle [-2,\;2]$$, as the equation $$\displaystyle |t| = 2$$ give two equal roots.

First, in order to have two distinct real roots, we must have
$$\displaystyle Δ = m^2−64(2m−15) > 0 ⇐⇒ (m−8)(m−120) > 0 ⇐⇒ m < 8, m>120.$$

Next, we note that $$\displaystyle f(2) = 16 · 4 − 15 > 0$$, so either $$\displaystyle t_1 < t_2 < −2$$ or $$\displaystyle 2 < t_1 < t_2$$. The first case cannot happen. Indeed, if it does then by the Viete formula $$\displaystyle \frac{m}{16}= t_1 + t_2 < −4 ⇐⇒ m < −64 =⇒ t_1t_2 =\frac{2m− 15}{16}<0$$, which is impossible.

Thus we get $$\displaystyle 2 < t_1 < t_2$$, and each of the two equations $$\displaystyle x+\frac{1}{x}=t_1$$, $$\displaystyle x+\frac{1}{x}=t_2$$ has two real positive distinct roots, which we denote by $$\displaystyle x_1,\; x'_1$$ and $$\displaystyle x_2,\; x'_2$$ respectively.

Note that $$\displaystyle x_1x'_1 = x_2x'_2 = 1.$$ We can assume that $$\displaystyle 1 < x_1 < x_2$$, which implies that $$\displaystyle 1 > x_1 > x_2$$. Then we have $$\displaystyle x'_2, x'_1, x_1, x_2$$ form an increasing geometric progression.

Therefore, $$\displaystyle x_2 = (x_1)^3, x_2 = (x_1)^3$$, which implies that

$$\displaystyle t_2 = x_2 +\frac {1}{x_2}=x_2+x'_2= (x_1)^3 + (x'_1)^3=(x_1+x'_1)((x_1)^2-x_1x'_1+(x'_1)^2)=(x_1+x'_1)((x_1+x'_1)^2-3x_1x'_1)=t_1((t_1)^2-3$$

Then $$\displaystyle \frac{m}{16}= t_1 + t_2=t_1((t_1)^2-2)$$, and hence $$\displaystyle m=16t_1((t_1)^2-2)=t_1(16(t_1)^2-32)=t_1 [(mt_1 − 2m+ 15) − 32] = m(t_1)^2 − (2m+ 17)t_1,$$

which gives $$\displaystyle m =\frac{17t_1}{(t_1)^2 − 2t_1 − 1}$$.

Substituting this value of m into the equation $$\displaystyle f(t1) = 16(t_1)^2−mt_1+2m− 15 = 0$$ we obtain
$$\displaystyle 16(t_1)^4 − 31(t_1)^3 − 48(t_1)^2 + 64t_1 + 15 = 0.$$

Denote $$\displaystyle y = 2t_1$$, we have $$\displaystyle y^4 − 4y^3 − 12y^2 + 32y + 15 = 0 ⇐⇒ (y − 5)(y + 3)(y2 − 2y − 1) = 0.$$

From this it follows that the unique possible value of $$\displaystyle y$$ for which $$\displaystyle t_1 > 2$$ is $$\displaystyle y = 5$$. Hence $$\displaystyle t_1 = \frac{5}{2}$$ and so $$\displaystyle m = 170$$.

Conversely, for $$\displaystyle m = 170$$ the equation $$\displaystyle 16x^4 − 170x^3 + 357x^2 − 170x +6 = 0$$ has four distinct roots $$\displaystyle \frac{1}{8}\;\frac{1}{2},\;2,\;8$$ which obviously form a geometric progression with the ratio $$\displaystyle r = 4$$.

Thus the only solution to the problem is $$\displaystyle m = 170$$.

Well-known member
16x^4−mx^3+(2m+17)x^2−mx+16=0 has four distinct roots forming a geometric progression.

so let the roots be a/r^3, a/r, ar and ar^3 (the common ratio = r^2)

product of roots = a^4 = 16/16 = 1

so the polynomial is q( x +1/r^3)(x+1/r)(x)(x+r)(x+ r^3)

or n( r^3 x + 1)(r x + 1) ( x + r) (x+r^3)

as the coefficients are symmetric we have n =1

now take the product to get r^4 = 16

r = 2 or -2(only possible values)

by putting 2 and then -2 we can get the result)