# Find the values of the parameter "q"

#### anemone

##### MHB POTW Director
Staff member
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.

$q=9$ 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.

#### Opalg

##### MHB Oldtimer
Staff member
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.

$q=9$ 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 $q=34.$

Staff member
Hi Opalg,