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(K)=1/2K-4
We move all terms to the left:
(K)-(1/2K-4)=0
Domain of the equation: 2K-4)!=0We get rid of parentheses
K∈R
K-1/2K+4=0
We multiply all the terms by the denominator
K*2K+4*2K-1=0
Wy multiply elements
2K^2+8K-1=0
a = 2; b = 8; c = -1;
Δ = b2-4ac
Δ = 82-4·2·(-1)
Δ = 72
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}$$K_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$$K_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-6\sqrt{2}}{2*2}=\frac{-8-6\sqrt{2}}{4} $$K_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+6\sqrt{2}}{2*2}=\frac{-8+6\sqrt{2}}{4} $
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