[equations involving radicals] 2(x2 + x + 1) + sqrt(x2+x+1) - 3 = 0

mxrie

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Hello again! I'm back with another problem from our prerequisite math class and as the title suggests, it's a problem with radicals and we have to find its solutions.

The given problem is: 2(x2 + x + 1) + sqrt(x2+x+1) - 3 = 0

So far what I have done is:
  1. sqrt(x2+x+1) = -2x2 + 2x + 1 [isolate the radical on one side]
  2. x2+x+1 = 4x4+8x3+4x+1 [square both sides to remove the radical]
  3. 4x4 + 8x3 - x2 + 3x = 0 [combine like terms]

And at this point, I find myself stuck because I don't really know...
  • if I did the correct process
  • and how I'm supposed to solve the fourth degree equation

Hoping someone can enlighten me with this process. Thank you in advance!
 
Hello again! I'm back with another problem from our prerequisite math class and as the title suggests, it's a problem with radicals and we have to find its solutions.

The given problem is: 2(x2 + x + 1) + sqrt(x2+x+1) - 3 = 0

So far what I have done is:
  1. sqrt(x2+x+1) = -2x2 + 2x + 1 [isolate the radical on one side]
  2. x2+x+1 = 4x4+8x3+4x+1 [square both sides to remove the radical]
  3. 4x4 + 8x3 - x2 + 3x = 0 [combine like terms]

And at this point, I find myself stuck because I don't really know...
  • if I did the correct process
  • and how I'm supposed to solve the fourth degree equation

Hoping someone can enlighten me with this process. Thank you in advance!

What you've done so far is standard and correct (except for one sign error, if I'm right); it just left you at a point where (apart from factoring out an x) you can't easily continue. There may be some tricks that would make it possible to factor the resulting cubic, but it would take time to do so, at the least. (Actually, you should try the rational zero theorem, which is the standard alternative way to solve a polynomial.)

If you don't get a solution this way, you back up and see if any side paths might take you in a better direction.

Did you notice that x2 + x + 1 appears in two places, and that is the only form in which x appears? When that happens, a substitution can be useful: set u = x2 + x + 1. You will be able to solve for u, and then use that to solve for x. Don't forget to check for extraneous solutions, of course.
 
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What you've done so far is standard and correct (except for one sign error, if I'm right); it just left you at a point where (apart from factoring out an x) you can't easily continue. There may be some tricks that would make it possible to factor the resulting cubic, but it would take time to do so, at the least. (Actually, you should try the rational zero theorem, which is the standard alternative way to solve a polynomial.)

If you don't get a solution this way, you back up and see if any side paths might take you in a better direction.

Did you notice that x2 + x + 1 appears in two places, and that is the only form in which x appears? When that happens, a substitution can be useful: set u = x2 + x + 1. You will be able to solve for u, and then use that to solve for x. Don't forget to check for extraneous solutions, of course.

I got it! I set sqrt(x2 + x + 1) as u then proceeded to factor and substitute - which ended up with 0 and 1 as solutions. Thank you for your help!
 
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