1/3z^2=9

Solution for 1/3z^2=9 equation:

1/3z^2=9

We move all terms to the left:

1/3z^2-(9)=0


Domain of the equation: 3z^2!=0

z^2!=0/3

z^2!=√0

z!=0

z∈R


We multiply all the terms by the denominator


-9*3z^2+1=0

Wy multiply elements

-27z^2+1=0

a = -27; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-27)·1
Δ = 108
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:

$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{3}}{2*-27}=\frac{0-6\sqrt{3}}{-54}
=-\frac{6\sqrt{3}}{-54}
=-\frac{\sqrt{3}}{-9}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{3}}{2*-27}=\frac{0+6\sqrt{3}}{-54}
=\frac{6\sqrt{3}}{-54}
=\frac{\sqrt{3}}{-9}
$