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√(2x-1)=-3 Help?

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by squaring both sides\[2x-1=9\]
\[\sqrt{2x - 1} = -3\] Square on both sides. \[2x - 1 = 9\] Quite simple now, can you do it?
solve for 'x' from here

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Other answers:

don't even start this problem \(\sqrt{2x-1}\) is positive, whereas \(-3\) is negative, so there is no solution
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
which is good example of making sure you think before you apply some method to a problem
Hmm... the sqrt(9) isn't -3, but I don't see how that disallows -3^2
i am assuming you are looking for a real value for \(x\) and not two complex numbers whose square is \(-3\)
but @satellite73 sir this gives the condition of the question but he needs to solve for 'x'
I didn't think from that point of view.
That was a good point.
Allright, I think I'm getting it. Thanks for all the help. :)
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 \[\sqrt{2\times 5-1}=\sqrt{9}=3\neq -3\]
u r right sir :) sorry!
Excellent answer, satellite73!
@math0101 for steps u can first solve it & prove this wrong or other answer is as @satellite73 sir's answer :)

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