4q+4=2q+4+32/q

Solution for 4q+4=2q+4+32/q equation:

4q+4=2q+4+32/q

We move all terms to the left:

4q+4-(2q+4+32/q)=0


Domain of the equation: q)!=0

q!=0/1

q!=0

q∈R


We add all the numbers together, and all the variables

4q-(2q+32/q+4)+4=0

We get rid of parentheses

4q-2q-32/q-4+4=0

We multiply all the terms by the denominator


4q*q-2q*q-4*q+4*q-32=0

We add all the numbers together, and all the variables

4q*q-2q*q-32=0

Wy multiply elements

4q^2-2q^2-32=0

We add all the numbers together, and all the variables

2q^2-32=0

a = 2; b = 0; c = -32;
Δ = b2-4ac
Δ = 02-4·2·(-32)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*2}=\frac{-16}{4}
=-4
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*2}=\frac{16}{4}
=4
$