9/2k-3=4/k=+1

Solution for 9/2k-3=4/k=+1 equation:

9/2k-3=4/k=+1

We move all terms to the left:

9/2k-3-(4/k)=0


Domain of the equation: 2k!=0

k!=0/2

k!=0

k∈R



Domain of the equation: k)!=0

k!=0/1

k!=0

k∈R


We add all the numbers together, and all the variables

9/2k-(+4/k)-3=0

We get rid of parentheses

9/2k-4/k-3=0

We calculate fractions


9k/2k^2+(-8k)/2k^2-3=0

We multiply all the terms by the denominator


9k+(-8k)-3*2k^2=0

Wy multiply elements

-6k^2+9k+(-8k)=0

We get rid of parentheses

-6k^2+9k-8k=0

We add all the numbers together, and all the variables

-6k^2+k=0

a = -6; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-6)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-6}=\frac{-2}{-12}
=1/6
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-6}=\frac{0}{-12}
=0
$