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anmolsd1995
I tried to solve second equation this way:

z^3-25z=0

z^3=25z

z^2=25

z= +/-5

Since z is not positive it must be -5 then.

can anyone please correct me where i am going wrong.

If its not positive it could have been that it is 0 or -ive, so by taking the value to RHS, you removed a case in which z could have been 0

z^3-25z=0
z (z^2-25) = 0

Now z can be 0 or z can be +/- 5

But then it cant be +ive, it can be 0 or -5, Since 0 is neither positive nor negative
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anmolsd1995
I tried to solve second equation this way:

z^3-25z=0

z^3=25z

z^2=25

z= +/-5

Since z is not positive it must be -5 then.

can anyone please correct me where i am going wrong.

Hey anmolsd1995 ,
  • If you cancel \(z\) and go from \(z^3=25z\) to \(z^2=25\), you will get rid of a case where z can be equal to zero.
  • In equality, we should cancel terms iff we know that it is non zero.
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