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math0101

  • 3 years ago

√(2x-1)=-3 Help?

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  1. jiteshmeghwal9
    • 3 years ago
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    by squaring both sides\[2x-1=9\]

  2. pratu043
    • 3 years ago
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    \[\sqrt{2x - 1} = -3\] Square on both sides. \[2x - 1 = 9\] Quite simple now, can you do it?

  3. jiteshmeghwal9
    • 3 years ago
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    solve for 'x' from here

  4. anonymous
    • 3 years ago
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    don't even start this problem \(\sqrt{2x-1}\) is positive, whereas \(-3\) is negative, so there is no solution

  5. anonymous
    • 3 years ago
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    don't square both sides, don't do anything at all a positive number cannot equal a negative one say "no solution" and be done

  6. anonymous
    • 3 years ago
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    which is good example of making sure you think before you apply some method to a problem

  7. math0101
    • 3 years ago
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    Hmm... the sqrt(9) isn't -3, but I don't see how that disallows -3^2

  8. anonymous
    • 3 years ago
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    i am assuming you are looking for a real value for \(x\) and not two complex numbers whose square is \(-3\)

  9. jiteshmeghwal9
    • 3 years ago
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    but @satellite73 sir this gives the condition of the question but he needs to solve for 'x'

  10. pratu043
    • 3 years ago
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    I didn't think from that point of view.

  11. pratu043
    • 3 years ago
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    That was a good point.

  12. math0101
    • 3 years ago
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    Allright, I think I'm getting it. Thanks for all the help. :)

  13. anonymous
    • 3 years ago
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    if so (and i really really doubt it) then this method \[2x-1=9\] \[2x=10\] \[x=5\] will certainly not work, and the check is easy \[\sqrt{2\times 5-1}=\sqrt{9}=3\neq -3\]

  14. jiteshmeghwal9
    • 3 years ago
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    hmmmm......

  15. jiteshmeghwal9
    • 3 years ago
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    u r right sir :) sorry!

  16. math0101
    • 3 years ago
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    Excellent answer, satellite73!

  17. jiteshmeghwal9
    • 3 years ago
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    @math0101 for steps u can first solve it & prove this wrong or other answer is as @satellite73 sir's answer :)

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