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1-1/5x^2=0
Domain of the equation: 5x^2!=0We multiply all the terms by the denominator
x^2!=0/5
x^2!=√0
x!=0
x∈R
1*5x^2-1=0
Wy multiply elements
5x^2-1=0
a = 5; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·5·(-1)
Δ = 20
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{5}}{2*5}=\frac{0-2\sqrt{5}}{10} =-\frac{2\sqrt{5}}{10} =-\frac{\sqrt{5}}{5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{5}}{2*5}=\frac{0+2\sqrt{5}}{10} =\frac{2\sqrt{5}}{10} =\frac{\sqrt{5}}{5} $
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