Solve for x x^4-41x^2+400=0

$x^{4} - 41 x^{2} + 400 = 0$

Step 1

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

$u^{2} - 41 u + 400 = 0$

$u = x^{2}$

Step 2

Factor $u^{2} - 41 u + 400$ 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 $400$ and whose sum is $- 41$ .

$- 25, - 16$

Write the factored form using these integers.

$(u - 25) (u - 16) = 0$

Step 3

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

$u - 25 = 0$

$u - 16 = 0$

Step 4

Set $u - 25$ equal to $0$ and solve for $u$ .

Set $u - 25$ equal to $0$ .

$u - 25 = 0$

Add $25$ to both sides of the equation.

$u = 25$

Step 5

Set $u - 16$ equal to $0$ and solve for $u$ .

Set $u - 16$ equal to $0$ .

$u - 16 = 0$

Add $16$ to both sides of the equation.

$u = 16$

Step 6

The final solution is all the values that make $(u - 25) (u - 16) = 0$ true.

$u = 25, 16$

Step 7

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

$x^{2} = 25$

${(x^{2})}^{1} = 16$

Step 8

Solve the first equation for $x$ .

$x^{2} = 25$

Step 9

Solve the equation for $x$ .

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

$x = \pm \sqrt{25}$

Simplify $\pm \sqrt{25}$ .

Rewrite $25$ as $5^{2}$ .

$x = \pm \sqrt{5^{2}}$

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

$x = \pm 5$

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.

$x = 5$

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

$x = - 5$

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

$x = 5, - 5$

Step 10

Solve the second equation for $x$ .

${(x^{2})}^{1} = 16$

Step 11

Solve the equation for $x$ .

Remove parentheses.

$x^{2} = 16$

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

$x = \pm \sqrt{16}$

Simplify $\pm \sqrt{16}$ .

Rewrite $16$ as $4^{2}$ .

$x = \pm \sqrt{4^{2}}$

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

$x = \pm 4$

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.

$x = 4$

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

$x = - 4$

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

$x = 4, - 4$

Step 12

The solution to $x^{4} - 41 x^{2} + 400 = 0$ is $x = 5, - 5, 4, - 4$ .

$x = 5, - 5, 4, - 4$

Solve for x x^4-41x^2+400=0

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