Convert to Set Notation x^4-45x^2+324<0

$x^{4} - 45 x^{2} + 324 < 0$

Solve $- 6 < x < - 3 or 3 < x < 6$ .

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

$u^{2} - 45 u + 324 = 0$

$u = x^{2}$

Factor $u^{2} - 45 u + 324$ 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 $324$ and whose sum is $- 45$ .

$- 36, - 9$

Write the factored form using these integers.

$(u - 36) (u - 9) = 0$

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

$u - 36 = 0$

$u - 9 = 0$

Set $u - 36$ equal to $0$ and solve for $u$ .

Set $u - 36$ equal to $0$ .

$u - 36 = 0$

Add $36$ to both sides of the equation.

$u = 36$

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.

$u = 9$

The final solution is all the values that make $(u - 36) (u - 9) = 0$ true.

$u = 36, 9$

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

$x^{2} = 36$

${(x^{2})}^{1} = 9$

Solve the first equation for $x$ .

$x^{2} = 36$

Solve the equation for $x$ .

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

$x = \pm \sqrt{36}$

Simplify $\pm \sqrt{36}$ .

Rewrite $36$ as $6^{2}$ .

$x = \pm \sqrt{6^{2}}$

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

$x = \pm 6$

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.

$x = 6$

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

$x = - 6$

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

$x = 6, - 6$

Solve the second equation for $x$ .

${(x^{2})}^{1} = 9$

Solve the equation for $x$ .

Remove parentheses.

$x^{2} = 9$

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.

$x = 3, - 3$

The solution to $x^{4} - 45 x^{2} + 324 = 0$ is $x = 6, - 6, 3, - 3$ .

$x = 6, - 6, 3, - 3$

Use each root to create test intervals.

$x < - 6$

$- 6 < x < - 3$

$- 3 < x < 3$

$3 < x < 6$

$x > 6$

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 $x < - 6$ to see if it makes the inequality true.

Choose a value on the interval $x < - 6$ and see if this value makes the original inequality true.

$x = - 8$

Replace $x$ with $- 8$ in the original inequality.

${(- 8)}^{4} - 45 {(- 8)}^{2} + 324 < 0$

The left side $1540$ is not less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 6 < x < - 3$ to see if it makes the inequality true.

Choose a value on the interval $- 6 < x < - 3$ and see if this value makes the original inequality true.

$x = - 4$

Replace $x$ with $- 4$ in the original inequality.

${(- 4)}^{4} - 45 {(- 4)}^{2} + 324 < 0$

The left side $- 140$ is less than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = 0$

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

${(0)}^{4} - 45 {(0)}^{2} + 324 < 0$

The left side $324$ is not less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $3 < x < 6$ to see if it makes the inequality true.

Choose a value on the interval $3 < x < 6$ and see if this value makes the original inequality true.

$x = 4$

Replace $x$ with $4$ in the original inequality.

${(4)}^{4} - 45 {(4)}^{2} + 324 < 0$

The left side $- 140$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 6$ to see if it makes the inequality true.

Choose a value on the interval $x > 6$ and see if this value makes the original inequality true.

$x = 8$

Replace $x$ with $8$ in the original inequality.

${(8)}^{4} - 45 {(8)}^{2} + 324 < 0$

The left side $1540$ is not less than the right side $0$ , which means that the given statement is false.

False

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

$x < - 6$ False

$- 6 < x < - 3$ True

$- 3 < x < 3$ False

$3 < x < 6$ True

$x > 6$ False

$x < - 6$ False

$- 6 < x < - 3$ True

$- 3 < x < 3$ False

$3 < x < 6$ True

$x > 6$ False

The solution consists of all of the true intervals.

$- 6 < x < - 3$ or $3 < x < 6$

Use the inequality $- 6 < x < - 3 or 3 < x < 6$ to build the set notation.

${x | - 6 < x < - 3 or 3 < x < 6}$

Convert to Set Notation x^4-45x^2+324<0

Related Questions

Support LCMGCF.com by disabling your adblocker.