Complex problem with additional equation

1. Aug 16, 2013

V0ODO0CH1LD

1. The problem statement, all variables and given/known data

Let $\lambda$ be the real solutions to the equation $\sqrt{\lambda +9}+\sqrt{2\lambda +17}=12$. Then what is the sum of the solutions $z$, with $Re\ z>0$, of the equation $z^4=\lambda -32$?

2. Relevant equations

3. The attempt at a solution

I tried substituting $\lambda =z^4+32$ into $\sqrt{\lambda +9}+\sqrt{2\lambda +17}=12$ to see if something simplified, but nothing did. Am I left with hacking this one out? Because I refuse to believe that I should try to actually solve $\sqrt{\lambda +9}+\sqrt{2\lambda +17}=12$ for $\lambda$.

Is there a "better" way?

2. Aug 16, 2013

haruspex

Perhaps there's a better way, but solving that really isn't hard at all.

3. Aug 16, 2013

QuantumCurt

Remember when you're solving equations involving radicals, you need to isolate an individual radical on one side of the equation. If you subtract one of the radical terms from both sides, you can then square both sides. Pay attention to the fact that you will be squaring a binomial on one side.

There might be an easier way, but not that I'm aware of. The only method I know of is slogging through the algebra.

Last edited: Aug 16, 2013
4. Aug 16, 2013

QuantumCurt

$$\sqrt{\lambda +9}+\sqrt{2\lambda +17}=12$$

As I mentioned above, you want to isolate one of the radicals, so we'll subtract one of them from both sides of the equation.

$$\sqrt{2\lambda +17}=12-\sqrt{\lambda +9}$$

Now we want to start trying to eliminate the radicals, which we can do by squaring both sides.

$$(\sqrt{2\lambda +17})^2=(12-\sqrt{\lambda +9})^2$$

Now go through and apply the exponents. Remember that $(\sqrt{a})^2=a$ and keep in mind that you're squaring a binomial on the right side. Afterwards, you'll need to eliminate the second radical.

There's a fair amount of algebra here, but solving for $\lambda$ isn't too terrible difficult.

5. Aug 17, 2013

V0ODO0CH1LD

It's not about being terribly difficult. I already solved it that way and got $\lambda = 16$, but the whole process was very "ugly" (not hard). This is one of those competition math problems, and the first one that I had to resort to hacking. Usually there is some identity or theorem that simplifies the problem in case it looks like it's gonna be very laborious and likely for you to make mistakes. But every now and again it seem like there is a problem that is very time consuming in these things.

Thanks!

6. Aug 17, 2013

tiny-tim

Hi V0ODO0CH1LD!
In that case, you can guess there must be a quick way!

Two things to notice immediately:

i] it's obviously an increasing function, and so can have only one solution

ii] 2λ + 17 must be odd, so both those squares must be odd​

Since you're expecting a quick way, the first pair to check would be 5 + 7 …

which of course works!

7. Aug 17, 2013

pasmith

I don't think the question actually requires that you find $\lambda$; you only need to find the sum of the roots in question as a function of $\lambda$.

We know that $\lambda$ is real, so $\lambda - 32$ is also real. So we need the solutions of $z^4 = C$ with $C \in \mathbb{R}$.

If $C > 0$, then $C = k^4$ for some $k > 0$. Then the solutions of $z^4 = k^4$ are $k$, $-k$ and $\pm ik$. The only one with strictly positive real part is $k$.

If $C < 0$, then $C = -k^4$ for some $k > 0$. Then either $z^2 = ik^2$ or $z^2 = -ik^2$. In the first case,
$$z = \pm \frac{k}{\sqrt{2}} (1 + i)$$
and in the second,
$$z = \pm \frac{k}{\sqrt{2}}(1 - i)$$
Thus there are a conjugate pair of solutions with strictly positive real part, and the sum of those solutions is $k\sqrt{2}$.

So all you need to know is whether $\lambda > 32$. Since $f(x) = \sqrt{x +9}+\sqrt{2x +17}$ is strictly increasing, this is equivalent to asking whether $f(\lambda) = 12 > f(32)$.

8. Aug 17, 2013

V0ODO0CH1LD

are you saying that I don't need to find $k$? Because the acceptable answer is $2\sqrt{2}$. Which requires me to find $k$.