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1/5+m(2/5)=1
We move all terms to the left:
1/5+m(2/5)-(1)=0
determiningTheFunctionDomain m(2/5)-1+1/5=0
We add all the numbers together, and all the variables
m(+2/5)-1+1/5=0
We multiply parentheses
2m^2-1+1/5=0
We multiply all the terms by the denominator
2m^2*5+1-1*5=0
We add all the numbers together, and all the variables
2m^2*5-4=0
Wy multiply elements
10m^2-4=0
a = 10; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·10·(-4)
Δ = 160
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{10}}{2*10}=\frac{0-4\sqrt{10}}{20} =-\frac{4\sqrt{10}}{20} =-\frac{\sqrt{10}}{5} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{10}}{2*10}=\frac{0+4\sqrt{10}}{20} =\frac{4\sqrt{10}}{20} =\frac{\sqrt{10}}{5} $
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