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some body help me plzzzz solve
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\[3+\sqrt{z-6}=\sqrt{2-9}\]
just checking, is the 2 on the RHS of the equation supposed to be a z?
well..you familiar with complex nos then madam ?
@ebbflo srry i wz wrong
yes it is Z
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\[3+\sqrt{z-6}=\sqrt{z+9}\]
this the right question
ok \[3+\sqrt{z-6}=\sqrt{z+9}\] \[(3+\sqrt{z-6})^2=(\sqrt{z+9})^2\] \[9+6\sqrt{z-6}+z-6=z+9\] \[6\sqrt{z-6}=6\] \[\sqrt{z-6}=1\] \[z-6=1\] \[z=7\]
z=7
err, i made a mistake, one sec, tried to save all the code I wrote for the first eq...
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no take it back, its good
yep..pretty good :)
u cool thanks a lot :D
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