Precalculus Examples

$9 x^{4} - 97 x^{2} + 144 \leq 0$

Step 1

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

$9 u^{2} - 97 u + 144 = 0$

$u = x^{2}$

Step 2

Factor by grouping.

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Step 2.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 9 \cdot 144 = 1296$ and whose sum is $b = - 97$ .

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Step 2.1.1

Factor $- 97$ out of $- 97 u$ .

$9 u^{2} - 97 u + 144 = 0$

Step 2.1.2

Rewrite $- 97$ as $- 16$ plus $- 81$

$9 u^{2} + (- 16 - 81) u + 144 = 0$

Step 2.1.3

Apply the distributive property.

$9 u^{2} - 16 u - 81 u + 144 = 0$

Step 2.2

Factor out the greatest common factor from each group.

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Step 2.2.1

Group the first two terms and the last two terms.

$(9 u^{2} - 16 u) - 81 u + 144 = 0$

Step 2.2.2

Factor out the greatest common factor (GCF) from each group.

$u (9 u - 16) - 9 (9 u - 16) = 0$

Step 2.3

Factor the polynomial by factoring out the greatest common factor, $9 u - 16$ .

$(9 u - 16) (u - 9) = 0$

Step 3

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

$9 u - 16 = 0$

$u - 9 = 0$

Step 4

Set $9 u - 16$ equal to $0$ and solve for $u$ .

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Step 4.1

Set $9 u - 16$ equal to $0$ .

$9 u - 16 = 0$

Step 4.2

Solve $9 u - 16 = 0$ for $u$ .

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Step 4.2.1

Add $16$ to both sides of the equation.

$9 u = 16$

Step 4.2.2

Divide each term in $9 u = 16$ by $9$ and simplify.

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Step 4.2.2.1

Divide each term in $9 u = 16$ by $9$ .

$\frac{9 u}{9} = \frac{16}{9}$

Step 4.2.2.2

Simplify the left side.

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Step 4.2.2.2.1

Cancel the common factor of $9$ .

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Step 4.2.2.2.1.1

Cancel the common factor.

$\frac{9 u}{9} = \frac{16}{9}$

Step 4.2.2.2.1.2

Divide $u$ by $1$ .

$u = \frac{16}{9}$

Step 5

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

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Step 5.1

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

$u - 9 = 0$

Step 5.2

Add $9$ to both sides of the equation.

$u = 9$

Step 6

The final solution is all the values that make $(9 u - 16) (u - 9) = 0$ true.

$u = \frac{16}{9}, 9$

Step 7

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

$x^{2} = \frac{16}{9}$

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

Step 8

Solve the first equation for $x$ .

$x^{2} = \frac{16}{9}$

Step 9

Solve the equation for $x$ .

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Step 9.1

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

$x = \pm \sqrt{\frac{16}{9}}$

Step 9.2

Simplify $\pm \sqrt{\frac{16}{9}}$ .

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Step 9.2.1

Rewrite $\sqrt{\frac{16}{9}}$ as $\frac{\sqrt{16}}{\sqrt{9}}$ .

$x = \pm \frac{\sqrt{16}}{\sqrt{9}}$

Step 9.2.2

Simplify the numerator.

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Step 9.2.2.1

Rewrite $16$ as $4^{2}$ .

$x = \pm \frac{\sqrt{4^{2}}}{\sqrt{9}}$

Step 9.2.2.2

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

$x = \pm \frac{4}{\sqrt{9}}$

Step 9.2.3

Simplify the denominator.

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Step 9.2.3.1

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

$x = \pm \frac{4}{\sqrt{3^{2}}}$

Step 9.2.3.2

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

$x = \pm \frac{4}{3}$

Step 9.3

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

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Step 9.3.1

First, use the positive value of the $\pm$ to find the first solution.

$x = \frac{4}{3}$

Step 9.3.2

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

$x = - \frac{4}{3}$

Step 9.3.3

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

$x = \frac{4}{3}, - \frac{4}{3}$

Step 10

Solve the second equation for $x$ .

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

Step 11

Solve the equation for $x$ .

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Step 11.1

Remove parentheses.

$x^{2} = 9$

Step 11.2

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

$x = \pm \sqrt{9}$

Step 11.3

Simplify $\pm \sqrt{9}$ .

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Step 11.3.1

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

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

Step 11.3.2

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

$x = \pm 3$

Step 11.4

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

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Step 11.4.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 3$

Step 11.4.2

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

$x = - 3$

Step 11.4.3

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

$x = 3, - 3$

Step 12

The solution to $9 x^{4} - 97 x^{2} + 144 = 0$ is $x = \frac{4}{3}, - \frac{4}{3}, 3, - 3$ .

$x = \frac{4}{3}, - \frac{4}{3}, 3, - 3$

Step 13

Use each root to create test intervals.

$x < - 3$

$- 3 < x < - \frac{4}{3}$

$- \frac{4}{3} < x < \frac{4}{3}$

$\frac{4}{3} < x < 3$

$x > 3$

Step 14

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

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Step 14.1

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

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Step 14.1.1

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

$x = - 6$

Step 14.1.2

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

$9 {(- 6)}^{4} - 97 {(- 6)}^{2} + 144 \leq 0$

Step 14.1.3

The left side $8316$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 14.2

Test a value on the interval $- 3 < x < - \frac{4}{3}$ to see if it makes the inequality true.

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Step 14.2.1

Choose a value on the interval $- 3 < x < - \frac{4}{3}$ and see if this value makes the original inequality true.

$x = - 2$

Step 14.2.2

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

$9 {(- 2)}^{4} - 97 {(- 2)}^{2} + 144 \leq 0$

Step 14.2.3

The left side $- 100$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 14.3

Test a value on the interval $- \frac{4}{3} < x < \frac{4}{3}$ to see if it makes the inequality true.

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Step 14.3.1

Choose a value on the interval $- \frac{4}{3} < x < \frac{4}{3}$ and see if this value makes the original inequality true.

$x = 0$

Step 14.3.2

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

$9 {(0)}^{4} - 97 {(0)}^{2} + 144 \leq 0$

Step 14.3.3

The left side $144$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 14.4

Test a value on the interval $\frac{4}{3} < x < 3$ to see if it makes the inequality true.

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Step 14.4.1

Choose a value on the interval $\frac{4}{3} < x < 3$ and see if this value makes the original inequality true.

$x = 2$

Step 14.4.2

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

$9 {(2)}^{4} - 97 {(2)}^{2} + 144 \leq 0$

Step 14.4.3

The left side $- 100$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 14.5

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

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Step 14.5.1

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

$x = 6$

Step 14.5.2

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

$9 {(6)}^{4} - 97 {(6)}^{2} + 144 \leq 0$

Step 14.5.3

The left side $8316$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 14.6

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

$x < - 3$ False

$- 3 < x < - \frac{4}{3}$ True

$- \frac{4}{3} < x < \frac{4}{3}$ False

$\frac{4}{3} < x < 3$ True

$x > 3$ False

$x < - 3$ False

$- 3 < x < - \frac{4}{3}$ True

$- \frac{4}{3} < x < \frac{4}{3}$ False

$\frac{4}{3} < x < 3$ True

$x > 3$ False

Step 15

The solution consists of all of the true intervals.

$- 3 \leq x \leq - \frac{4}{3}$ or $\frac{4}{3} \leq x \leq 3$

Step 16

Convert the inequality to interval notation.

$[- 3, - \frac{4}{3}] \cup [\frac{4}{3}, 3]$

Step 17