
#1
May412, 10:14 AM

P: 55

3z1+2√z=32
This equation gets solved when we assume the z as any variable's^{2} and turn that into a quadratic form. Is there any other way of solving this equation? 



#2
May412, 10:29 AM

P: 606

Well, when we square stuff, which ammounts to basically the same thing: [itex]3z33=2\sqrt{z}\Longrightarrow 9z^2198z+1089=4z\Longrightarrow 9z^2202z+1089=0\Longrightarrow z_1=13.444\,,\,\,z_2=9[/itex] . As many times with these exercises, only the second number above is a solution to the original equation. DonAntonio 



#3
May412, 11:21 AM

P: 55

But, the assumption thing looks simple enough. 


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