1 Answer | Add Yours
This equation is reducible to quadratic in form. All we need to do is to check the variable from the first term and to notice that is the square of the variable from the second term.
We'll apply substitution technique to solve this equation.
Let w^2 = t => w^4 = t^2
We'll re-write the equation in t:
t^2 - 6t - 2 = 0
We'll apply quadratic formula:
t1 = [6+sqrt(36 + 8)]/2
t1 = (6+2sqrt11)/2
t1 = 3+sqrt11
t2 = 3 - sqrt11
But w^2 = t1 => w^2 = 3 + sqrt11 => w1 = +sqrt(3+sqrt11)
w2 = -sqrt(3+sqrt11)
w^2 = t2 => w^2 = 3 - sqrt11
We notice that sqrt11 = 3.31 > 3 => 3 - sqrt11 < 0.
Since it is not allowed to have negative radicand, then we cannot take the square root of 3 - sqrt11.
Therefore, the only real solutions of the given equation are `+-` sqrt(3+sqrt11).
We’ve answered 319,175 questions. We can answer yours, too.Ask a question