4x+3/2x=7/2x

Solution for 4x+3/2x=7/2x equation:

4x+3/2x=7/2x

We move all terms to the left:

4x+3/2x-(7/2x)=0


Domain of the equation: 2x!=0

x!=0/2

x!=0

x∈R



Domain of the equation: 2x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

4x+3/2x-(+7/2x)=0

We get rid of parentheses

4x+3/2x-7/2x=0

We multiply all the terms by the denominator


4x*2x+3-7=0

We add all the numbers together, and all the variables

4x*2x-4=0

Wy multiply elements

8x^2-4=0

a = 8; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·8·(-4)
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{2}}{2*8}=\frac{0-8\sqrt{2}}{16}
=-\frac{8\sqrt{2}}{16}
=-\frac{\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{2}}{2*8}=\frac{0+8\sqrt{2}}{16}
=\frac{8\sqrt{2}}{16}
=\frac{\sqrt{2}}{2}
$