Finding the unknowns of irrational equation

In summary, this homework question is supposedly solvable, but you need to use trial and error to find the correct equations.
  • #1
asd1249jf

Homework Statement


For the following irrational equation

[tex]x^2 + 7x + 10 + \sqrt{x^2 + 7x + 12} = 0[/tex]

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.

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

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?
 
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  • #2
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
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
Dick said:
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
 
Last edited by a moderator:
  • #5
And yes, this can be solved very easily.

Set [tex]u = x^2 + 7x + 10[/tex]

[tex]u + \sqrt{u + 2} = 0[/tex]

[tex]\sqrt{u + 2} = -u[/tex]

[tex]u + 2 = u^2 [/tex]

[tex]u^2 - u - 2 = 0[/tex]

u = 2, -1

Now set the each corresponding u to two different equations

[tex]x^2 + 7x + 10 = 2[/tex]

[tex]x^2 + 7x + 10 = -1[/tex]

Solve for x.
 
  • #6
You are a freaking GENIUS! Thanks a lot!
 
  • #7
Doh!
 
  • #8
Remember to get rid of false solutions!

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

1. What is an irrational equation?

An irrational equation is an equation where the unknown variable is under a radical sign, such as square root, cube root, or any other nth root. These equations cannot be solved using traditional algebraic methods and may have infinite solutions.

2. How do you find the unknowns in an irrational equation?

The first step is to isolate the radical term on one side of the equation. Then, you can raise both sides of the equation to the power that will eliminate the radical. Repeat this process until the radical disappears. Finally, solve for the unknown variable using traditional algebraic methods.

3. Can irrational equations have more than one solution?

Yes, irrational equations can have multiple solutions. This is because raising both sides of the equation to a power can result in both positive and negative solutions, and some equations may have infinite solutions.

4. Are there any special cases when solving an irrational equation?

Yes, there are some special cases to consider when solving an irrational equation. For example, if the radical term is squared, you must check your solutions to ensure they are valid for the original equation. Additionally, if the equation contains more than one radical term, you may need to simplify before solving.

5. Can technology be used to solve irrational equations?

Yes, technology such as graphing calculators or computer programs can be used to solve irrational equations. These tools can quickly and accurately find solutions, but it is still important to understand the steps and concepts behind solving these equations by hand.

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