Solve for x x^4-5x^2=-6

$x^{4} - 5 x^{2} = - 6$

Step 1

Add $6$ to both sides of the equation.

$x^{4} - 5 x^{2} + 6 = 0$

Step 2

Factor the left side of the equation.

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

${(x^{2})}^{2} - 5 x^{2} + 6 = 0$

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$u^{2} - 5 u + 6 = 0$

Factor $u^{2} - 5 u + 6$ 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 $6$ and whose sum is $- 5$ .

$- 3, - 2$

Write the factored form using these integers.

$(u - 3) (u - 2) = 0$

Replace all occurrences of $u$ with $x^{2}$ .

$(x^{2} - 3) (x^{2} - 2) = 0$

Step 3

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

$x^{2} - 3 = 0$

$x^{2} - 2 = 0$

Step 4

Set $x^{2} - 3$ equal to $0$ and solve for $x$ .

Set $x^{2} - 3$ equal to $0$ .

$x^{2} - 3 = 0$

Solve $x^{2} - 3 = 0$ for $x$ .

Add $3$ to both sides of the equation.

$x^{2} = 3$

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

$x = \pm \sqrt{3}$

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.

$x = \sqrt{3}$

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

$x = - \sqrt{3}$

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

$x = \sqrt{3}, - \sqrt{3}$

Step 5

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

Set $x^{2} - 2$ equal to $0$ .

$x^{2} - 2 = 0$

Solve $x^{2} - 2 = 0$ for $x$ .

Add $2$ to both sides of the equation.

$x^{2} = 2$

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

$x = \pm \sqrt{2}$

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.

$x = \sqrt{2}$

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

$x = - \sqrt{2}$

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

$x = \sqrt{2}, - \sqrt{2}$

Step 6

The final solution is all the values that make $(x^{2} - 3) (x^{2} - 2) = 0$ true.

$x = \sqrt{3}, - \sqrt{3}, \sqrt{2}, - \sqrt{2}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = \sqrt{3}, - \sqrt{3}, \sqrt{2}, - \sqrt{2}$

Decimal Form:

$x = 1.73205080 \dots, - 1.73205080 \dots, 1.41421356 \dots, - 1.41421356 \dots$

Solve for x x^4-5x^2=-6

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