(2x)^-3/2x=0

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Solution for (2x)^-3/2x=0 equation:



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

The end solution:
$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*4}=\frac{0-4\sqrt{3}}{8} =-\frac{4\sqrt{3}}{8} =-\frac{\sqrt{3}}{2} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*4}=\frac{0+4\sqrt{3}}{8} =\frac{4\sqrt{3}}{8} =\frac{\sqrt{3}}{2} $

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