MHB Find the values of the parameter "q"

[anemone]

Hi MHB,

Problem:

Find the sum of all positive real values of the parameter $q$ for which the equation $(x^2+4x+4+136)^2=16q(5x^2+4x+136)$ has exactly three distinct real solutions.

Answer:
$q=9$ and $q=34$.

I am aware the question is actually asking us to find all possible values of $q$ when the given quartic equation has one repeated real root, and two other distinct real roots.

Attempt:

The original equation $(x^2+4x+4q+136)^2=16q(5x^2+4x+136)$ can be rewritten as

$(x^2+4x+136+4q)^2=16q(4x^2+x^2+4x+136)$

If I let $k=x^2+4x+136$, the equation above becomes

$(k+4q)^2=16q(4x^2+k)$

Expanding both sides of the equation and then rearranging them as a quadratic equation in terms of $k$, we have

$k^2+8qk+16q^2=64qx^2+16qk$

$k^2-8qk-64qx^2+16q^2=0$

$(k-4q)^2-16q^2-64qx^2+16q^2=0$

$(k-4q)^2-64qx^2=0$

$(k-4q-8x\sqrt{q})(k-4q+8x\sqrt{q})=0$

Replacing the expression for $k$ back into the equation above, we get

$(x^2+4x+136-4q-8x\sqrt{q})(x^2+4x+136-4q+8x\sqrt{q})=0$

And in order to satisfy the requirement set by the problem and to find all possible values of $q$, we can do so by equating the discriminants of the quadratic factors, one after another and we should expect two $q$ values to be found.

Hence, by equating the discriminant for the first factor $x^2+4x+136-4q-8x\sqrt{q}=0$ to zero, we obtain:

$(4-8\sqrt{q})^2-4(136-4q)=0$

$\sqrt{q}=3$ or $\sqrt{q}=-2.2$ and since $\sqrt{q}>0$, we get $q=9$ as one of the solutions.

Now, if we set the discriminant of the second factor as zero, we get:

$(4+8\sqrt{q})^2-4(136-4q)=0$

$\sqrt{q}=-3$ or $\sqrt{q}=2.2$ and since $\sqrt{q}>0$, we get $q=4.84$ as the other solution.

I checked the answer when $q=2.2$ and we would only get all 4 complex roots if $q=2.2$ but with $q=9$, everything is fine as is.

My question is, I don't know how to find the answer where $q=34$...any ideas?

By the way, I know there must be other ways to approach this problem, so if you don't wish to go through my silly attempt, I welcome you to post another method to solve the problem and I will appreciate whatever help that you are going to offer me.

Thanks in advance.

 

[Opalg]

Problem:

Find the sum of all positive real values of the parameter $q$ for which the equation $(x^2+4x+4+136)^2=16q(5x^2+4x+136)$ has exactly three distinct real solutions.

Answer:
$q=9$ and $q=34$.

$\vdots$

... we get

$(x^2+4x+136-4q-8x\sqrt{q})(x^2+4x+136-4q+8x\sqrt{q})=0$

And in order to satisfy the requirement set by the problem and to find all possible values of $q$, we can do so by equating the discriminants of the quadratic factors, one after another and we should expect two $q$ values to be found.

Hence, by equating the discriminant for the first factor $x^2+4x+136-4q-8x\sqrt{q}=0$ to zero, we obtain:

$(4-8\sqrt{q})^2-4(136-4q)=0$

$\sqrt{q}=3$ or $\sqrt{q}=-2.2$ and since $\sqrt{q}>0$, we get $q=9$ as one of the solutions.

Now, if we set the discriminant of the second factor as zero, we get:

$(4+8\sqrt{q})^2-4(136-4q)=0$

$\sqrt{q}=-3$ or $\sqrt{q}=2.2$ and since $\sqrt{q}>0$, we get $q=4.84$ as the other solution.

I checked the answer when $q=2.2$ and we would only get all 4 complex roots if $q=2.2$ but with $q=9$, everything is fine as is.

My question is, I don't know how to find the answer where $q=34$...any ideas?

I had to spend over four hours in trains today, going to and from a meeting in London, and I spent all that time struggling with this problem. I got close to a solution, but your approach is far neater than mine.

The missing solution comes this way: In the equation $(x^2+4x+136-4q-8x\sqrt{q})(x^2+4x+136-4q+8x\sqrt{q})=0,$ you looked at the case where the repeated root comes in the first of the two factors, and when it comes in the second factor, but you overlooked the possibility that it might come once in each factor. That can only happen when the two factors are equal, and that in turn can only happen when $8x\sqrt{q}=0.$ That implies $x=0$, and the two factors then both reduce to $136 - 4q=0$, or $q=34.$

 

[anemone]

Hi Opalg,

Thank you so much for your reply.

Ah! I missed checking the likelihood that the repeated root might occur once in each factor and I wouldn't have thought of this if you hadn't pointed it out for me. I am so grateful that I can always depend on MHB to find for the missing part of the "math puzzle"!

When I read that you spent all the time sitting in the train struggling with this problem, it melts my heart so thoroughly. I hope there is a day where I can buy you a coffee (or tea, whatever that you would like to) or deliver one to your house, a cup of hot coffee that is specifically just for you! :D