Radical Equation

Date: 04/12/97 at 06:01:03
From: bob
Subject: Radical Equation

I'm helping my cousin on a practice exam and I am stumped on this 
problem.

   3x + (5x^(1/2)) - 28 = 0.

It is driving me crazy.  Thanks for your help.

Sincerely,
Bob

Date: 04/12/97 at 16:07:33
From: Doctor Mason
Subject: Re: Radical Equation

Hi, Bob.

Your equation involves a 1/2 power. This makes it a radical equation 
rather than a polynomial equation.

To solve it, I would first isolate the term with the 1/2 exponent 
thus:

   5x^(1/2) = 28 - 3x.      

Since  (a^(1/2))^2 = a^1 by property of exponents, our next step will 
be to square both sides, taking care to square the entire side as a 
unit and NOT term by term.

   (5x^(1/2))^2 = (28 - 3x)^2     
      5^2 * x^1 = (28-3x)(28-3x)   
            25x = 784 - 168x + 9x^2.     

Since the equation is quadratic we will set it equal to zero.
   
              0 = 9x^2 - 193x + 784.        

Then we factor.
  
              0 = (9x - 49)(x - 16),       

which will give us roots of 49/9 and 16.

Since we raised the power of this equation when solving it, we must be 
sure to check both solutions as one or both could be extraneous roots, 
i.e. solutions to the quadratic equation, but not to the original 
equation.

Indeed, when we check 16 we find that 3(16) + 5(16)^1/2 - 28 = 40  and 
not 0.  But, when checking 49/9, we happily see that the equation is 
true.
  
   3(49/9) + 5(49/9)^1/2 - 28 = 49/3  +  35/3  - 28  = 0.

Hope this helps! 

-Doctor Mason,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/