Question 319350
w^4 - 2w^2 - 2 = 0
:
Find w^2 using the quadratic formula
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
in this equation
x=w^2, a=1, b=-2, c=-2
{{{w^2 = (-(-2) +- sqrt(-2^2-4*1*-2 ))/(2*1) }}}
:
{{{w^2 = (2 +- sqrt(4 - (-8) ))/2 }}}
:
Two solutions
{{{w^2 = (2 + sqrt(12 ))/2 }}}
{{{w^2 = (2 + sqrt(4*3))/2 }}}
extract the sqrt of 4
{{{w^2 = (2 + 2*sqrt(3))/2 }}}
Cancel the 2's
{{{w^2 = 1 + sqrt(3) }}}
Find the square root of both sides
w = {{{sqrt(1+sqrt(3))}}}; a good solution
:
but we also have
{{{w^2 = (2 - sqrt(12 ))/2 }}}
{{{w^2 = (2 - sqrt(4*3))/2 }}}
extract the sqrt of 4
{{{w^2 = (2 - 2*sqrt(3))/2 }}}
Cancel the 2's
{{{w^2 = 1 - sqrt(3) }}}
Find the square root of both sides
w = {{{sqrt(1-sqrt(3))}}}
However sqrt of 3 > 1, therefore we have square root of a neg number
:
I'm going to say the exact solution using radicals
w = +{{{sqrt(1+sqrt(3))}}}, -{{{sqrt(1+sqrt(3))}}}
:
Confirm this with a graph of this equation
{{{ graph( 300, 200, -4, 4, -4, 4, x^4-2x^2-2) }}}