1/r+3=r-5/r

Solution for 1/r+3=r-5/r equation:

1/r+3=r-5/r

We move all terms to the left:

1/r+3-(r-5/r)=0


Domain of the equation: r!=0

r∈R



Domain of the equation: r)!=0

r!=0/1

r!=0

r∈R


We add all the numbers together, and all the variables

1/r-(+r-5/r)+3=0

We get rid of parentheses

1/r-r+5/r+3=0

We multiply all the terms by the denominator


-r*r+3*r+1+5=0

We add all the numbers together, and all the variables

3r-r*r+6=0

Wy multiply elements

-1r^2+3r+6=0

a = -1; b = 3; c = +6;
Δ = b2-4ac
Δ = 32-4·(-1)·6
Δ = 33
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{33}}{2*-1}=\frac{-3-\sqrt{33}}{-2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{33}}{2*-1}=\frac{-3+\sqrt{33}}{-2}
$