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- Feb 14, 2012

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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.