1/5x^2=25

Solution for 1/5x^2=25 equation:

1/5x^2=25

We move all terms to the left:

1/5x^2-(25)=0


Domain of the equation: 5x^2!=0

x^2!=0/5

x^2!=√0

x!=0

x∈R


We multiply all the terms by the denominator


-25*5x^2+1=0

Wy multiply elements

-125x^2+1=0

a = -125; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-125)·1
Δ = 500
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:

$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{5}}{2*-125}=\frac{0-10\sqrt{5}}{-250}
=-\frac{10\sqrt{5}}{-250}
=-\frac{\sqrt{5}}{-25}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{5}}{2*-125}=\frac{0+10\sqrt{5}}{-250}
=\frac{10\sqrt{5}}{-250}
=\frac{\sqrt{5}}{-25}
$