1/4x^2+1=17

Solution for 1/4x^2+1=17 equation:

1/4x^2+1=17

We move all terms to the left:

1/4x^2+1-(17)=0


Domain of the equation: 4x^2!=0

x^2!=0/4

x^2!=√0

x!=0

x∈R


We add all the numbers together, and all the variables

1/4x^2-16=0

We multiply all the terms by the denominator


-16*4x^2+1=0

Wy multiply elements

-64x^2+1=0

a = -64; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-64)·1
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*-64}=\frac{-16}{-128}
=1/8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*-64}=\frac{16}{-128}
=-1/8
$