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11/z-9/2z=1
We move all terms to the left:
11/z-9/2z-(1)=0
Domain of the equation: z!=0
z∈R
Domain of the equation: 2z!=0We calculate fractions
z!=0/2
z!=0
z∈R
22z/2z^2+(-9z)/2z^2-1=0
We multiply all the terms by the denominator
22z+(-9z)-1*2z^2=0
Wy multiply elements
-2z^2+22z+(-9z)=0
We get rid of parentheses
-2z^2+22z-9z=0
We add all the numbers together, and all the variables
-2z^2+13z=0
a = -2; b = 13; c = 0;
Δ = b2-4ac
Δ = 132-4·(-2)·0
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{169}=13$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-13}{2*-2}=\frac{-26}{-4} =6+1/2 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+13}{2*-2}=\frac{0}{-4} =0 $
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