Solve for z z^4-37z^2+36=0

$z^{4} - 37 z^{2} + 36 = 0$

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

$u^{2} - 37 u + 36 = 0$

$u = z^{2}$

Factor $u^{2} - 37 u + 36$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $36$ and whose sum is $- 37$ .

$- 36, - 1$

Write the factored form using these integers.

$(u - 36) (u - 1) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$u - 36 = 0$

$u - 1 = 0$

Set $u - 36$ equal to $0$ and solve for $u$ .

Set $u - 36$ equal to $0$ .

$u - 36 = 0$

Add $36$ to both sides of the equation.

$u = 36$

Set $u - 1$ equal to $0$ and solve for $u$ .

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

$u - 1 = 0$

Add $1$ to both sides of the equation.

$u = 1$

The final solution is all the values that make $(u - 36) (u - 1) = 0$ true.

$u = 36, 1$

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

$z^{2} = 36$

${(z^{2})}^{1} = 1$

Solve the first equation for $z$ .

$z^{2} = 36$

Solve the equation for $z$ .

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

$z = \pm \sqrt{36}$

Simplify $\pm \sqrt{36}$ .

Rewrite $36$ as $6^{2}$ .

$z = \pm \sqrt{6^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$z = \pm 6$

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

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

$z = 6$

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

$z = - 6$

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

$z = 6, - 6$

Solve the second equation for $z$ .

${(z^{2})}^{1} = 1$

Solve the equation for $z$ .

Remove parentheses.

$z^{2} = 1$

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

$z = \pm \sqrt{1}$

Any root of $1$ is $1$ .

$z = \pm 1$

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

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

$z = 1$

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

$z = - 1$

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

$z = 1, - 1$

The solution to $z^{4} - 37 z^{2} + 36 = 0$ is $z = 6, - 6, 1, - 1$ .

$z = 6, - 6, 1, - 1$

Solve for z z^4-37z^2+36=0

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