# Homework Help: Finding the unknowns of irrational equation

1. Oct 3, 2007

### l46kok

1. The problem statement, all variables and given/known data
For the following irrational equation

$$x^2 + 7x + 10 + \sqrt{x^2 + 7x + 12} = 0$$

Find all possible unknown of X.

2. Relevant equations

None. Just your ability to solve equations.

3. The attempt at a solution

First of all, I am not allowed to use a calculator to solve this (During an exam).

Second of all, I am not sure why this is considered as irrational equation, but I went ahead and attempted to solve it.

$$x^2+7x+10 = -\sqrt{x^2+7x+12}$$
$$x^2+7x+10 = -\sqrt{(x+4)(x+3)}$$
$$(x^2+7x+10)^2 = (x+4)(x+3)$$
$$x^4+14x^3+140x + 69x^2 + 100 = x^2 + 7x + 12$$
$$x^4 + 14x^3 + 68x^2 + 133x + 88 = 0$$

And I am completely stuck here. I know no method which you can solve this fourth order equation by hand (Remember, NO CALCULATORS TO SOLVE THIS)

Any ideas?

2. Oct 3, 2007

### Dick

The only thing you can try to do is to factor. And it only can be factored into two quadratics. And it would take a considerable amount of trial and error to find that. Do you want me to say that it's a really poor exam question? Because I would agree with that.

3. Oct 3, 2007

### Avodyne

I don't know any way to do it either. I wonder if there's a misprint; if the 10 was a 12, it would be much easier ...

4. Oct 4, 2007

### l46kok

Unfortunately, it actually is an exam question (Supposedly solvable). This was a question from Korean National Standarized Test. From what I heard, there's a real simpler way of doing this, but I just cannot figure out what it is.

http://img385.imageshack.us/img385/5272/12qu6.jpg [Broken]

Last edited by a moderator: May 3, 2017
5. Oct 4, 2007

### shuh

And yes, this can be solved very easily.

Set $$u = x^2 + 7x + 10$$

$$u + \sqrt{u + 2} = 0$$

$$\sqrt{u + 2} = -u$$

$$u + 2 = u^2$$

$$u^2 - u - 2 = 0$$

u = 2, -1

Now set the each corresponding u to two different equations

$$x^2 + 7x + 10 = 2$$

$$x^2 + 7x + 10 = -1$$

Solve for x.

6. Oct 4, 2007

### l46kok

You are a freaking GENIUS! Thanks a lot!!!

7. Oct 4, 2007

Doh!

8. Oct 4, 2007

### arildno

Remember to get rid of false solutions!

u=2 cannot be used, since
$$2+\sqrt{2+2}=4\neq{0}$$