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2^x+1/2^x-1=2^2
We move all terms to the left:
2^x+1/2^x-1-(2^2)=0
Domain of the equation: 2^x!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
2^x+1/2^x-5=0
We multiply all the terms by the denominator
2^x*2^x-5*2^x+1=0
Wy multiply elements
4x^2-10x+1=0
a = 4; b = -10; c = +1;
Δ = b2-4ac
Δ = -102-4·4·1
Δ = 84
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{21}}{2*4}=\frac{10-2\sqrt{21}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{21}}{2*4}=\frac{10+2\sqrt{21}}{8} $
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