Question 442413
1. Let x = w^2
Then we can write the equation as
x^2 - 12x - 2 = 0
Since the factorization is not obvious, solve using the quadratic formula:
{{{x = (12 +- sqrt(12^2 + 8))/2}}}
{{{x = (12 +- sqrt(152))/2}}}
152 can be factored as 4*38, so we can simplify the radical:
{{{x = (12 +- 2sqrt(38))/2}}}
This simplifies to {{{x = 6 +- sqrt(38)}}}
But {{{x = w^2}}}, so {{{w = sqrt(x)}}}
{{{w = sqrt(6 +- sqrt(38))}}},{{{-sqrt(6 +- sqrt(38))}}}
If we are restricting ourselves to real numbers, we can't take the square root
of a negative number so the two solutions are:
{{{w = sqrt(6 + sqrt(38))}}},{{{-sqrt(6 + sqrt(38))}}} [approximately +-3.5]
This is a bit messy, but as a check of our result we can graph the function and look for the zero crossings.
We see that the function crosses the x-axis around x = -3.5 and x = 3.5.  
{{{graph(400,400,-5,5,-40,40,x^4-12x^2-2)}}}
2. If the two roots are {{{-sqrt(2)}}},{{{7*sqrt(2)}}}, then the equation can be factored as:
{{{(x+sqrt(2))(x-7sqrt(2)) = 0}}}
Multiply using FOIL and collect terms:
{{{x^2 -6sqrt(2)x - 14 = 0}}}