Precalculus Examples

$9 x^{4} + 44 x^{2} - 5$

Step 1

Set $9 x^{4} + 44 x^{2} - 5$ equal to $0$ .

$9 x^{4} + 44 x^{2} - 5 = 0$

Step 2

Solve for $x$ .

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Step 2.1

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

$9 u^{2} + 44 u - 5 = 0$

$u = x^{2}$

Step 2.2

Factor by grouping.

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Step 2.2.1

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 $a \cdot c = 9 \cdot - 5 = - 45$ and whose sum is $b = 44$ .

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Step 2.2.1.1

Factor $44$ out of $44 u$ .

$9 u^{2} + 44 (u) - 5 = 0$

Step 2.2.1.2

Rewrite $44$ as $- 1$ plus $45$

$9 u^{2} + (- 1 + 45) u - 5 = 0$

Step 2.2.1.3

Apply the distributive property.

$9 u^{2} - 1 u + 45 u - 5 = 0$

Step 2.2.2

Factor out the greatest common factor from each group.

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Step 2.2.2.1

Group the first two terms and the last two terms.

$(9 u^{2} - 1 u) + 45 u - 5 = 0$

Step 2.2.2.2

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

$u (9 u - 1) + 5 (9 u - 1) = 0$

Step 2.2.3

Factor the polynomial by factoring out the greatest common factor, $9 u - 1$ .

$(9 u - 1) (u + 5) = 0$

Step 2.3

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

$9 u - 1 = 0$

$u + 5 = 0$

Step 2.4

Set $9 u - 1$ equal to $0$ and solve for $u$ .

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Step 2.4.1

Set $9 u - 1$ equal to $0$ .

$9 u - 1 = 0$

Step 2.4.2

Solve $9 u - 1 = 0$ for $u$ .

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Step 2.4.2.1

Add $1$ to both sides of the equation.

$9 u = 1$

Step 2.4.2.2

Divide each term in $9 u = 1$ by $9$ and simplify.

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Step 2.4.2.2.1

Divide each term in $9 u = 1$ by $9$ .

$\frac{9 u}{9} = \frac{1}{9}$

Step 2.4.2.2.2

Simplify the left side.

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Step 2.4.2.2.2.1

Cancel the common factor of $9$ .

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Step 2.4.2.2.2.1.1

Cancel the common factor.

$\frac{9 u}{9} = \frac{1}{9}$

Step 2.4.2.2.2.1.2

Divide $u$ by $1$ .

$u = \frac{1}{9}$

Step 2.5

Set $u + 5$ equal to $0$ and solve for $u$ .

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Step 2.5.1

Set $u + 5$ equal to $0$ .

$u + 5 = 0$

Step 2.5.2

Subtract $5$ from both sides of the equation.

$u = - 5$

Step 2.6

The final solution is all the values that make $(9 u - 1) (u + 5) = 0$ true.

$u = \frac{1}{9}, - 5$

Step 2.7

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

$x^{2} = \frac{1}{9}$

${(x^{2})}^{1} = - 5$

Step 2.8

Solve the first equation for $x$ .

$x^{2} = \frac{1}{9}$

Step 2.9

Solve the equation for $x$ .

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Step 2.9.1

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

$x = \pm \sqrt{\frac{1}{9}}$

Step 2.9.2

Simplify $\pm \sqrt{\frac{1}{9}}$ .

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Step 2.9.2.1

Rewrite $\sqrt{\frac{1}{9}}$ as $\frac{\sqrt{1}}{\sqrt{9}}$ .

$x = \pm \frac{\sqrt{1}}{\sqrt{9}}$

Step 2.9.2.2

Any root of $1$ is $1$ .

$x = \pm \frac{1}{\sqrt{9}}$

Step 2.9.2.3

Simplify the denominator.

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Step 2.9.2.3.1

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

$x = \pm \frac{1}{\sqrt{3^{2}}}$

Step 2.9.2.3.2

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

$x = \pm \frac{1}{3}$

Step 2.9.3

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

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Step 2.9.3.1

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

$x = \frac{1}{3}$

Step 2.9.3.2

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

$x = - \frac{1}{3}$

Step 2.9.3.3

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

$x = \frac{1}{3}, - \frac{1}{3}$

Step 2.10

Solve the second equation for $x$ .

${(x^{2})}^{1} = - 5$

Step 2.11

Solve the equation for $x$ .

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Step 2.11.1

Remove parentheses.

$x^{2} = - 5$

Step 2.11.2

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

$x = \pm \sqrt{- 5}$

Step 2.11.3

Simplify $\pm \sqrt{- 5}$ .

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Step 2.11.3.1

Rewrite $- 5$ as $- 1 (5)$ .

$x = \pm \sqrt{- 1 (5)}$

Step 2.11.3.2

Rewrite $\sqrt{- 1 (5)}$ as $\sqrt{- 1} \cdot \sqrt{5}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{5}$

Step 2.11.3.3

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i \sqrt{5}$

Step 2.11.4

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

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Step 2.11.4.1

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

$x = i \sqrt{5}$

Step 2.11.4.2

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

$x = - i \sqrt{5}$

Step 2.11.4.3

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

$x = i \sqrt{5}, - i \sqrt{5}$

Step 2.12

The solution to $9 x^{4} + 44 x^{2} - 5 = 0$ is $x = \frac{1}{3}, - \frac{1}{3}, i \sqrt{5}, - i \sqrt{5}$ .

$x = \frac{1}{3}, - \frac{1}{3}, i \sqrt{5}, - i \sqrt{5}$

Step 3