3z-1+2√z=32This equation gets solved when we assume the z as

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The equation 3z - 1 + 2√z = 32 can be transformed into a quadratic form by assuming z as a variable squared. Squaring both sides leads to the equation 9z^2 - 202z + 1089 = 0, which yields two potential solutions: z1 = 13.444 and z2 = 9. However, only z2 = 9 satisfies the original equation. The discussion emphasizes the effectiveness of this method while questioning if alternative solutions exist.
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3z-1+2√z=32
This equation gets solved when we assume the z as any variable's2 and turn that into a quadratic form.
Is there any other way of solving this equation?
 
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Kartik. said:
3z-1+2√z=32
This equation gets solved when we assume the z as any variable's2 and turn that into a quadratic form.
Is there any other way of solving this equation?
Well, when we square stuff, which ammounts to basically the same thing:

3z-33=-2\sqrt{z}\Longrightarrow 9z^2-198z+1089=4z\Longrightarrow 9z^2-202z+1089=0\Longrightarrow z_1=13.444\,,\,\,z_2=9 .

As many times with these exercises, only the second number above is a solution to the original equation.

DonAntonio
 
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DonAntonio said:
Well, when we square stuff, which ammounts to basically the same thing:

3z-33=-2\sqrt{z}\Longrightarrow 9z^2-198z+1089=4z\Longrightarrow 9z^2-202z+1089=0\Longrightarrow z_1=13.444\,,\,\,z_2=9 .

As many times with these exercises, only the second number above is a solution to the original equation.

DonAntonio
Thanks :D
But, the assumption thing looks simple enough.
 
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