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2/3z=8z=
We move all terms to the left:
2/3z-(8z)=0
Domain of the equation: 3z!=0We add all the numbers together, and all the variables
z!=0/3
z!=0
z∈R
-8z+2/3z=0
We multiply all the terms by the denominator
-8z*3z+2=0
Wy multiply elements
-24z^2+2=0
a = -24; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-24)·2
Δ = 192
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{3}}{2*-24}=\frac{0-8\sqrt{3}}{-48} =-\frac{8\sqrt{3}}{-48} =-\frac{\sqrt{3}}{-6} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*-24}=\frac{0+8\sqrt{3}}{-48} =\frac{8\sqrt{3}}{-48} =\frac{\sqrt{3}}{-6} $
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