Finding the unknowns of irrational equation

[asd1249jf]

Homework Statement

For the following irrational equation

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

Find all possible unknown of X.

Homework Equations

None. Just your ability to solve equations.

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?

 

[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.

 

[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 ...

 

[asd1249jf]

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.

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

 

[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.

 

[asd1249jf]

You are a freaking GENIUS! Thanks a lot!

 

[Dick]

Doh!

 

[arildno]

Remember to get rid of false solutions!

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