Convert to Set Notation w^4-50w^2+49<=0

$w^{4} - 50 w^{2} + 49 \leq 0$

Step 1

Solve $- 7 \leq w \leq - 1 or 1 \leq w \leq 7$ .

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

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

$u = w^{2}$

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.

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

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

$u - 49 = 0$

$u - 1 = 0$

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.

$u = 49$

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

$u = 49, 1$

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

$w^{2} = 49$

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

Solve the first equation for $w$ .

$w^{2} = 49$

Solve the equation for $w$ .

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

$w = \pm \sqrt{49}$

Simplify $\pm \sqrt{49}$ .

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

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

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

$w = \pm 7$

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.

$w = 7$

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

$w = - 7$

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

$w = 7, - 7$

Solve the second equation for $w$ .

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

Solve the equation for $w$ .

Remove parentheses.

$w^{2} = 1$

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

$w = \pm \sqrt{1}$

Any root of $1$ is $1$ .

$w = \pm 1$

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.

$w = 1$

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

$w = - 1$

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

$w = 1, - 1$

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

$w = 7, - 7, 1, - 1$

Use each root to create test intervals.

$w < - 7$

$- 7 < w < - 1$

$- 1 < w < 1$

$1 < w < 7$

$w > 7$

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 $w < - 7$ to see if it makes the inequality true.

Choose a value on the interval $w < - 7$ and see if this value makes the original inequality true.

$w = - 10$

Replace $w$ with $- 10$ in the original inequality.

${(- 10)}^{4} - 50 {(- 10)}^{2} + 49 \leq 0$

The left side $5049$ is greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 7 < w < - 1$ to see if it makes the inequality true.

Choose a value on the interval $- 7 < w < - 1$ and see if this value makes the original inequality true.

$w = - 4$

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

${(- 4)}^{4} - 50 {(- 4)}^{2} + 49 \leq 0$

The left side $- 495$ is less than the right side $0$ , which means that the given statement is always true.

True

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

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

$w = 0$

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

${(0)}^{4} - 50 {(0)}^{2} + 49 \leq 0$

The left side $49$ is greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $1 < w < 7$ to see if it makes the inequality true.

Choose a value on the interval $1 < w < 7$ and see if this value makes the original inequality true.

$w = 4$

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

${(4)}^{4} - 50 {(4)}^{2} + 49 \leq 0$

The left side $- 495$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $w > 7$ to see if it makes the inequality true.

Choose a value on the interval $w > 7$ and see if this value makes the original inequality true.

$w = 10$

Replace $w$ with $10$ in the original inequality.

${(10)}^{4} - 50 {(10)}^{2} + 49 \leq 0$

The left side $5049$ is greater 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.

$w < - 7$ False

$- 7 < w < - 1$ True

$- 1 < w < 1$ False

$1 < w < 7$ True

$w > 7$ False

$w < - 7$ False

$- 7 < w < - 1$ True

$- 1 < w < 1$ False

$1 < w < 7$ True

$w > 7$ False

The solution consists of all of the true intervals.

$- 7 \leq w \leq - 1$ or $1 \leq w \leq 7$

Step 2

Use the inequality $- 7 \leq w \leq - 1 or 1 \leq w \leq 7$ to build the set notation.

${w | - 7 \leq w \leq - 1 or 1 \leq w \leq 7}$

Step 3

Convert to Set Notation w^4-50w^2+49<=0

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