Expert: Paul Klarreich - 11/1/2008

Question
QUESTION: I am assuming that y+y^1/2-7 is classed as non polynomial.

The question was Solve y + y^(1/2) - 7 = 0

Can this be solved using the quadratic formula: (-b+/- root(b^2-4ac))/2a or because of the y^1/2 does it require a different approach.

the answer was asked for in surd form  C +/- root(5)    where C is a constant to be found.

I made it 4 +/- root(5) but that was with the formula -b+/- etc

Any clarity on this matter will be appreciated.
 
rg

ANSWER: Questioner:   rg
Category:  Advanced Math
Private:  No
 
Subject:  technique for solving polynomials vs non-polynomial
Question:  I am assuming that y+y^1/2-7 is classed as non polynomial.

The question was Solve y + y^(1/2) - 7 = 0

Can this be solved using the quadratic formula: (-b+/- root(b^2-4ac))/2a or because of the y^1/2 does it require a different approach.

the answer was asked for in surd form  C +/- root(5)    where C is a constant to be found.

I made it 4 +/- root(5) but that was with the formula -b+/- etc

Any clarity on this matter will be appreciated.

rg
...........................
Hi, rg,  

Your equation:

y + y^(1/2) - 7 = 0

can be rewritten using  x = sqrt(y) into:

x^2 + x - 7 = 0.

Now you solve using the Q.F.:
   - 1 +- sqrt(1 + 28)
x = -------------------
         2

   - 1 +- sqrt(29)
x = ---------------
         2

These are not the roots for y, of course, but:

y = x^2, so your roots would be:

   (- 1 +- sqrt(29))^2
y = --------------------
         4

   1 + 29 +- 2 sqrt(29)
y = ------------------
         4

   30 +- 2 sqrt(29)
y = ----------------
         4

   15 +- sqrt(29)
y = ----------------
         2

Check:

15 +- sqrt(29)   -1 -+ sqrt(29)
-------------- + -------------
      2          2
15 - 1
------ = 7, check
  2

I don't see how you got your sqrt(5) stuff in there.  Send my your work and I'll see what's going on.

(And nobody calls a + b sqrt()  a surd any more.  That word was obsolete when I was learning math.)


---------- FOLLOW-UP ----------

QUESTION: my apologies: I must have typed in the wrong question last night..........the one, (that is the question), that they said the answer to, should be written in c+- sqrt(5) was in fact :

y- 8y^(1/2)+11 = 0

To this I got 4+-sqrt(5) with the formula and without substitution ...which seems now incorrect.

And with substitution (y=x^2) like you said above ...  it seems that would  become ( 4 +- sqrt(5) )^2
         = 16 + 5 +-8sqrt(5)
         = 21 +- 8sqrt(5)

 hope that checks out and if so

thank you kindly for the response.

regards rg  

Answer
Yes, that checks out.

The (quadratic) form will be:

x^2 - 8x + 11 = 0

   8 +- sqrt(64-44)
x = -----------------
        2

   8 +- sqrt(20)
x = --------------
        2

   8 +- 2 sqrt(5)
x = -------------
        2

x = 4 +- sqrt(5)


This is the solution for x = sqrt(y), so

y = x^2 = (4 +- sqrt(5))^2, which you can easily compute.