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1/2k+2=+5/6k
We move all terms to the left:
1/2k+2-(+5/6k)=0
Domain of the equation: 2k!=0
k!=0/2
k!=0
k∈R
Domain of the equation: 6k)!=0We get rid of parentheses
k!=0/1
k!=0
k∈R
1/2k-5/6k+2=0
We calculate fractions
6k/12k^2+(-10k)/12k^2+2=0
We multiply all the terms by the denominator
6k+(-10k)+2*12k^2=0
Wy multiply elements
24k^2+6k+(-10k)=0
We get rid of parentheses
24k^2+6k-10k=0
We add all the numbers together, and all the variables
24k^2-4k=0
a = 24; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·24·0
Δ = 16
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{16}=4$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*24}=\frac{0}{48} =0 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*24}=\frac{8}{48} =1/6 $
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