(1/2k+1)=k

Solution for (1/2k+1)=k equation:

(1/2k+1)=k

We move all terms to the left:

(1/2k+1)-(k)=0


Domain of the equation: 2k+1)!=0

k∈R


We add all the numbers together, and all the variables

-1k+(1/2k+1)=0

We get rid of parentheses

-1k+1/2k+1=0

We multiply all the terms by the denominator


-1k*2k+1*2k+1=0

Wy multiply elements

-2k^2+2k+1=0

a = -2; b = 2; c = +1;
Δ = b2-4ac
Δ = 22-4·(-2)·1
Δ = 12
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{3}}{2*-2}=\frac{-2-2\sqrt{3}}{-4}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{3}}{2*-2}=\frac{-2+2\sqrt{3}}{-4}
$