Solve for y y^4-2y^2+1=0

$y^{4} - 2 y^{2} + 1 = 0$

Substitute $u = y^{2}$ into the equation. This will make the quadratic formula easy to use.

$u^{2} - 2 u + 1 = 0$

$u = y^{2}$

Factor using the perfect square rule.

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Rewrite $1$ as $1^{2}$ .

$u^{2} - 2 u + 1^{2} = 0$

Check the middle term by multiplying $2 a b$ and compare this result with the middle term in the original expression.

$2 a b = 2 \cdot u \cdot - 1$

Simplify.

$2 a b = - 2 u$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = u$ and $b = - 1$ .

${(u - 1)}^{2} = 0$

Set the $u - 1$ equal to $0$ .

$u - 1 = 0$

Add $1$ to both sides of the equation.

$u = 1$

Substitute the real value of $u = y^{2}$ back into the solved equation.

$y^{2} = 1$

Solve the equation for $y$ .

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Take the square root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt{1}$

The complete solution is the result of both the positive and negative portions of the solution.

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Any root of $1$ is $1$ .

$y = \pm 1$

The complete solution is the result of both the positive and negative portions of the solution.

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First, use the positive value of the $\pm$ to find the first solution.

$y = 1$

Next, use the negative value of the $\pm$ to find the second solution.

$y = - 1$

The complete solution is the result of both the positive and negative portions of the solution.

$y = 1, - 1$

Solve for y y^4-2y^2+1=0

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