Explanation

to solve for x in the equation

$x^2$ $+$ $\frac {2}{r^2}x$ $+ \frac {1}{r^4} = 0$ , we multiply throughout by the lowest common multiple (LCM) of the denominations to eliminate them...

LCM of the equation is $r^4$

so,

$r^4 \times x^2 + r^4 \times \frac{2}{r^2}$$x$ $+ r^4 \times \frac {1}{r^4}$ $= 0 \times r^4$

$r^4x^2$ $+ 2r^2x + 1 = 0$

also, this equation can be written as

$(r^2x)^2 + 2r^2x + 1 = 0$

now represent $r^2x$ by a letter, let use $b$

we can now rewrite the equation as

$b^2 + 2b + 1 = 0$

$b^2 + b + b + 1 = 0$

$b(b + 1) + 1(b + 1) = 0$

$(b+1)(b+1) = 0$

recall that $b = r^2x$

$r^2x + 1 = 0$

$r^2x = -1$

$x = \frac {-1}{r^2}$

to solve for x in the equation

$x^2$ $+$ $\frac {2}{r^2}x$ $+ \frac {1}{r^4} = 0$ , we multiply throughout by the lowest common multiple (LCM) of the denominations to eliminate them...

LCM of the equation is $r^4$

so,

$r^4 \times x^2 + r^4 \times \frac{2}{r^2}$$x$ $+ r^4 \times \frac {1}{r^4}$ $= 0 \times r^4$

$r^4x^2$ $+ 2r^2x + 1 = 0$

also, this equation can be written as

$(r^2x)^2 + 2r^2x + 1 = 0$

now represent $r^2x$ by a letter, let use $b$

we can now rewrite the equation as

$b^2 + 2b + 1 = 0$

$b^2 + b + b + 1 = 0$

$b(b + 1) + 1(b + 1) = 0$

$(b+1)(b+1) = 0$

recall that $b = r^2x$

$r^2x + 1 = 0$

$r^2x = -1$

$x = \frac {-1}{r^2}$