Well, when we square stuff, which ammounts to basically the same thing: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?
Thanks :DDonAntonio said:Well, when we square stuff, which ammounts to basically the same thing:
[itex]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[/itex] .
As many times with these exercises, only the second number above is a solution to the original equation.
DonAntonio
The value of z in this equation cannot be determined without more information. The equation can be simplified to 3z+2√z=33, which is a quadratic equation. To solve for z, we would need to know the value of either 3z or 2√z.
To solve this equation for z, you would need to manipulate the equation to isolate z on one side. This could involve factoring, completing the square, or using the quadratic formula. However, as mentioned in the previous answer, the equation cannot be solved without more information.
The degree of this equation is 2. This is because the highest power of z in the equation is 2, as seen in the term 2√z.
Yes, this equation can be solved using algebra. However, as mentioned before, more information is needed to find the value of z.
No, it is not possible to solve this equation without assuming a value for z. As mentioned before, the equation is a quadratic equation and therefore requires a known value to find the value of z.