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Boxman61

  • 4 years ago

Solve: ___ ______ √x-2 -1 = √2(x-3)

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  1. MCronin
    • 4 years ago
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    square both sides to get (x-2)-2\[\sqrt{x-2}\]+1=2x-6 x-1-2\[\sqrt{x-2}/]=2x-6 2\[\sqrt{x-2}\]=-x+5 \[\sqrt{x-2}\]=-x+5/2 square both sides again to get x-2=(-x+5/2)^2 x=(-x+5/2)^2+2 x=(x^2-10x+25)/4 +2 4x=x^2-10x+27 x^2-14x+27=0 quadratic from there

  2. Boxman61
    • 4 years ago
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    \[\sqrt{x-2} -1 = \sqrt{2(x-3)}\] \[(\sqrt{x-2})^{2} - (1)^{2} = \sqrt{2x-6}\] \[x-2-1 = 2x-6 \] \[x-3 = 2x-6\] \[-x = -3\] x=3

  3. MCronin
    • 4 years ago
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    no you have o square te entire leftside so its (sqrt(x-2)-1)^2 so you get -2sqrt(x-2)+(x-2)+1=-2sqrt(x-2)+x-1

  4. Boxman61
    • 4 years ago
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    hmm ok, my instructors answer is 3...trying to figure it out by your method

  5. MCronin
    • 4 years ago
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    problem is that you cant square the terms individualy to clear the sqrt

  6. Denebel
    • 4 years ago
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    Hmm.. it is x^2-14x-33=0

  7. MCronin
    • 4 years ago
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    doing itnot in my head at 2 am i got x^2-8x+12=0 but still not geting to "3"

  8. Boxman61
    • 4 years ago
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    i appreciate the help

  9. MCronin
    • 4 years ago
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    holy hell dropped the 2 on the right side gimme one sec

  10. MCronin
    • 4 years ago
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    @denebel has the correct quadratic and it also has one answer that is 3

  11. Denebel
    • 4 years ago
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    x+5-2*sqrt (x-2) = 2x x+5 = 2x + 2*sqrt (x-2) 5 = 2x + 2*sqrt (x-2) - x 5-x = 2*sqrt (x-2) (5-x)^2 = (2*sqrt (x-2))^2 x^2 - 10x + 25 = 4x - 8 0 = -x^2 + 14x + 33 0 = x^2 - 14x - 33

  12. Denebel
    • 4 years ago
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    x=3, x=11. Then you check each value by plugging it back into the original equation. Only x=3 checks out, therefore, x=3 is your answer.

  13. Denebel
    • 4 years ago
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    Oops Typo 0 = -x^2 + 14x - 33 0 = x^2 - 14x + 33

  14. Boxman61
    • 4 years ago
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    your awesome, thank you... now to break it down so i learn it lol

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