Solve for x x^4-50x^2+49=0

$x^{4} - 50 x^{2} + 49 = 0$

Step 1

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

$u^{2} - 50 u + 49 = 0$

$u = x^{2}$

Step 2

Factor $u^{2} - 50 u + 49$ 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 $49$ and whose sum is $- 50$ .

$- 49, - 1$

Write the factored form using these integers.

$(u - 49) (u - 1) = 0$

Step 3

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

$u - 49 = 0$

$u - 1 = 0$

Step 4

Set $u - 49$ equal to $0$ and solve for $u$ .

Set $u - 49$ equal to $0$ .

$u - 49 = 0$

Add $49$ to both sides of the equation.

$u = 49$

Step 5

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.

$u = 1$

Step 6

The final solution is all the values that make $(u - 49) (u - 1) = 0$ true.

$u = 49, 1$

Step 7

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

$x^{2} = 49$

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

Step 8

Solve the first equation for $x$ .

$x^{2} = 49$

Step 9

Solve the equation for $x$ .

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

$x = \pm \sqrt{49}$

Simplify $\pm \sqrt{49}$ .

Rewrite $49$ as $7^{2}$ .

$x = \pm \sqrt{7^{2}}$

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

$x = \pm 7$

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.

$x = 7$

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

$x = - 7$

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

$x = 7, - 7$

Step 10

Solve the second equation for $x$ .

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

Step 11

Solve the equation for $x$ .

Remove parentheses.

$x^{2} = 1$

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

$x = \pm \sqrt{1}$

Any root of $1$ is $1$ .

$x = \pm 1$

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.

$x = 1$

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

$x = - 1$

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

$x = 1, - 1$

Step 12

The solution to $x^{4} - 50 x^{2} + 49 = 0$ is $x = 7, - 7, 1, - 1$ .

$x = 7, - 7, 1, - 1$

Solve for x x^4-50x^2+49=0

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