Problem in solving this radical equation

[gamaz321]

I could go through most of the steps for this equation. However, I do not have much idea how to solve the value of t here. Once I find t it will be straightforward to find x. Requesting some help on this part of the process. Thank you.

 

[Deleted member 4993]

I could go through most of the steps for this equation. However, I do not have much idea how to solve the value of t here. Once I find t it will be straightforward to find x. Requesting some help on this part of the process. Thank you.

Do you know

Rational Root theorem?

 

[gamaz321]

I knew it. I had to brush up a bit. However, using the rational root theorem the possible factors are 1, -1, 3, and -3, and none of those are satisfying this equation.

 

[Dr.Peterson]

I knew it. I had to brush up a bit. However, using the rational root theorem the possible factors are 1, -1, 3, and -3, and none of those are satisfying this equation.

Or 9 or -9, but that doesn't help. I graphed the equation just to see whether the solution is anything nice, and it doesn't look at all rational. This seems like a rather hard problem.

Can you tell us more about it? If it is associated with a class, what methods have you learned? If not, where did it come from?

And can you show us an image of the actual problem, in case you have miscopied something, or missed some detail that could be a clue?

 

[blamocur]

I got the answers by pure cheating:
[math]x = 6\sqrt{3}+9[/math][math]t = 2\sqrt{3} + 3 = \frac{x}{3} = \sqrt{2x+3}[/math]Can someone on this forum figure out the way to arrive at this answers?

 

[blamocur]

I got the answers by pure cheating:
[math]x = 6\sqrt{3}+9[/math][math]t = 2\sqrt{3} + 3 = \frac{x}{3} = \sqrt{2x+3}[/math]Can someone on this forum figure out the way to arrive at this answers?

Now it bothers me that 'cheat' is an anagram of 'teach' -- is there a higher meaning to this discovery?

 

[lookagain]

gamaz321, a possible hint with a possible method is that you can set it up as

\(\displaystyle t^4 - 7t^3 + 21t + 9 \ = \ (t^2 + at - 3)(t^2 + bt - 3). \)

That will lead to a system of two unknowns with two equations.

 

[gamaz321]

Thanks, for the help. However, I never thought it would go this complicated. I am going to work on the hints though.
Regards.