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Find all real solutions

  1. Mar 22, 2006 #1
    I got this algebra problem that I have been trying to get around but the answer I get is always different from the book.

    It says: Find all the real solutions, if any of each equation.

    21). [tex]\sqrt{2x-3} +x=3[/tex].

    I tried to resolve it using the following method:

    2x-3+x^2=9
    (2x-12)+(x^2-9)=0
    2(x-6)+(x+3)(x-3)=0
    (x+3) and (x-3) will cancel each other so;
    2x-12=0
    x=12/2
    x=6.

    However, this is wrong and I know it, cause the answer does not fit the equation. So can anyone help me point what's wrong with what I did?
     
  2. jcsd
  3. Mar 22, 2006 #2

    HallsofIvy

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    Yes, that's wrong! (a+ b)2 is not a2+ b2 so [itex](\sqrt{2x-3}+x)^2[/itex] is not 2x-3+ x!

    I recommend you subtract x from both sides of th equation to get
    [itex]\sqrt{2x-3}= 3- x[/itex], with only the square root on one side and then square both sides.
     
  4. Mar 22, 2006 #3
    If I use [itex]\sqrt{2x-3}= 3- x[/itex], I then worked it out to
    2x-3=9-6x+x^2
    2x+6x-x^2=9+3
    8x-x^2-12=0
    So I re-wrote as;
    -x^2+8x-12=0
    (x+6)=0 or (x+2)=0

    but then the answers add up as x=-6 or x=-2. and that can't be right..
     
  5. Mar 22, 2006 #4
    You didn't quite factor it correctly.
     
  6. Mar 22, 2006 #5
    Try factoring -x^2+8x-12=0 again. You may want to move everything to the right hand side to make it a bit easier.
     
  7. Mar 22, 2006 #6
    O.k, this time I tried the copleting the square method:

    -x^2+8x-12=0
    -x^2+8x=12
    (x+4)^2=12+16
    [itex](x+4)=\sqrt28[/itex]
    [itex]4+\sqrt28[/itex] or [itex]4-\sqrt28[/itex]
     
  8. Mar 22, 2006 #7
    The answers vary between -1.29 and 9.29, Still off
     
  9. Mar 22, 2006 #8
    -x^2 does not factor as (x+...)(x+...). That's your mistake.
     
  10. Mar 22, 2006 #9

    HallsofIvy

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    Rewrite -x2+ 8x- 12= 0 as x2- 8x+ 12= 0 by multiplying both sides by -1 (or "move everything to the right hand side" as was suggested before.

    x2- 8x+ 12= (x- ?)(x- ?).

    Of course, be sure to check solutions to the polynomial equation in the original equation!
     
  11. Mar 22, 2006 #10
    ah, I got it. You end up with (x- 6)(x- 2).
    and only x=2 would apply for the equation.
    Thanks
     
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