Solve the Inequality for x x^4-10x^2+9>=0

$x^{4} - 10 x^{2} + 9 \geq 0$

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

$u^{2} - 10 u + 9 = 0$

$u = x^{2}$

Factor $u^{2} - 10 u + 9$ 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 $9$ and whose sum is $- 10$ .

$- 9, - 1$

Write the factored form using these integers.

$(u - 9) (u - 1) = 0$

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

$u - 9 = 0$

$u - 1 = 0$

Set $u - 9$ equal to $0$ and solve for $u$ .

Set $u - 9$ equal to $0$ .

$u - 9 = 0$

Add $9$ to both sides of the equation.

$u = 9$

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 $(u - 9) (u - 1) = 0$ true.

$u = 9, 1$

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

$x^{2} = 9$

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

Solve the first equation for $x$ .

$x^{2} = 9$

Solve the equation for $x$ .

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

$x = \pm \sqrt{9}$

Simplify $\pm \sqrt{9}$ .

Rewrite $9$ as $3^{2}$ .

$x = \pm \sqrt{3^{2}}$

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

$x = \pm 3$

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.

$x = 3$

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

$x = - 3$

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

$x = 3, - 3$

Solve the second equation for $x$ .

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

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.

$x = 1, - 1$

The solution to $x^{4} - 10 x^{2} + 9 = 0$ is $x = 3, - 3, 1, - 1$ .

$x = 3, - 3, 1, - 1$

Use each root to create test intervals.

$x < - 3$

$- 3 < x < - 1$

$- 1 < x < 1$

$1 < x < 3$

$x > 3$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < - 3$ to see if it makes the inequality true.

Choose a value on the interval $x < - 3$ and see if this value makes the original inequality true.

$x = - 6$

Replace $x$ with $- 6$ in the original inequality.

${(- 6)}^{4} - 10 {(- 6)}^{2} + 9 \geq 0$

The left side $945$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 3 < x < - 1$ to see if it makes the inequality true.

Choose a value on the interval $- 3 < x < - 1$ and see if this value makes the original inequality true.

$x = - 2$

Replace $x$ with $- 2$ in the original inequality.

${(- 2)}^{4} - 10 {(- 2)}^{2} + 9 \geq 0$

The left side $- 15$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 1 < x < 1$ to see if it makes the inequality true.

Choose a value on the interval $- 1 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

${(0)}^{4} - 10 {(0)}^{2} + 9 \geq 0$

The left side $9$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $1 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $1 < x < 3$ and see if this value makes the original inequality true.

$x = 2$

Replace $x$ with $2$ in the original inequality.

${(2)}^{4} - 10 {(2)}^{2} + 9 \geq 0$

The left side $- 15$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 3$ to see if it makes the inequality true.

Choose a value on the interval $x > 3$ and see if this value makes the original inequality true.

$x = 6$

Replace $x$ with $6$ in the original inequality.

${(6)}^{4} - 10 {(6)}^{2} + 9 \geq 0$

The left side $945$ is greater than the right side $0$ , which means that the given statement is always true.

True

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 3$ True

$- 3 < x < - 1$ False

$- 1 < x < 1$ True

$1 < x < 3$ False

$x > 3$ True

$x < - 3$ True

$- 3 < x < - 1$ False

$- 1 < x < 1$ True

$1 < x < 3$ False

$x > 3$ True

The solution consists of all of the true intervals.

$x \leq - 3$ or $- 1 \leq x \leq 1$ or $x \geq 3$

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 3 or - 1 \leq x \leq 1 or x \geq 3$

Interval Notation:

$(- \infty, - 3] \cup [- 1, 1] \cup [3, \infty)$

Solve the Inequality for x x^4-10x^2+9>=0

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