Solve for x x^8+48x^4=49

$x^{8} + 48 x^{4} = 49$

Subtract $49$ from both sides of the equation.

$x^{8} + 48 x^{4} - 49 = 0$

Factor the left side of the equation.

Rewrite $x^{8}$ as ${(x^{4})}^{2}$ .

${(x^{4})}^{2} + 48 x^{4} - 49 = 0$

Let $u = x^{4}$ . Substitute $u$ for all occurrences of $x^{4}$ .

$u^{2} + 48 u - 49 = 0$

Factor $u^{2} + 48 u - 49$ 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 $48$ .

$- 1, 49$

Write the factored form using these integers.

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

Replace all occurrences of $u$ with $x^{4}$ .

$(x^{4} - 1) (x^{4} + 49) = 0$

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

$({(x^{2})}^{2} - 1) (x^{4} + 49) = 0$

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

$({(x^{2})}^{2} - 1^{2}) (x^{4} + 49) = 0$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x^{2}$ and $b = 1$ .

$(x^{2} + 1) (x^{2} - 1) (x^{4} + 49) = 0$

Simplify.

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

$(x^{2} + 1) (x^{2} - 1^{2}) (x^{4} + 49) = 0$

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$(x^{2} + 1) ((x + 1) (x - 1)) (x^{4} + 49) = 0$

Remove unnecessary parentheses.

$(x^{2} + 1) (x + 1) (x - 1) (x^{4} + 49) = 0$

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

$x^{2} + 1 = 0$

$x + 1 = 0$

$x - 1 = 0$

$x^{4} + 49 = 0$

Set $x^{2} + 1$ equal to $0$ and solve for $x$ .

Set $x^{2} + 1$ equal to $0$ .

$x^{2} + 1 = 0$

Solve $x^{2} + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

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

$x = \pm \sqrt{- 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i$

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.

$x = i$

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

$x = - i$

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

$x = i, - i$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set $x^{4} + 49$ equal to $0$ and solve for $x$ .

Set $x^{4} + 49$ equal to $0$ .

$x^{4} + 49 = 0$

Solve $x^{4} + 49 = 0$ for $x$ .

Subtract $49$ from both sides of the equation.

$x^{4} = - 49$

Take the 4th root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt[4]{- 49}$

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.

$x = \sqrt[4]{- 49}$

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

$x = - \sqrt[4]{- 49}$

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

$x = \sqrt[4]{- 49}, - \sqrt[4]{- 49}$

The final solution is all the values that make $(x^{2} + 1) (x + 1) (x - 1) (x^{4} + 49) = 0$ true.

$x = i, - i, - 1, 1, \sqrt[4]{- 49}, - \sqrt[4]{- 49}$

Solve for x x^8+48x^4=49

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