What are real solutions of equation w^4 -6w^2-2=0?

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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).

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