Trigonometry Examples

$(x^{4} + 5 x^{2} - 36) (2 x^{2} + 9 x - 5) = 0$

Step 1

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

$x^{4} + 5 x^{2} - 36 = 0$

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

Step 2

Set $x^{4} + 5 x^{2} - 36$ equal to $0$ and solve for $x$ .

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Step 2.1

Set $x^{4} + 5 x^{2} - 36$ equal to $0$ .

$x^{4} + 5 x^{2} - 36 = 0$

Step 2.2

Solve $x^{4} + 5 x^{2} - 36 = 0$ for $x$ .

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Step 2.2.1

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

$u^{2} + 5 u - 36 = 0$

$u = x^{2}$

Step 2.2.2

Factor $u^{2} + 5 u - 36$ using the AC method.

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Step 2.2.2.1

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 $- 36$ and whose sum is $5$ .

$- 4, 9$

Step 2.2.2.2

Write the factored form using these integers.

$(u - 4) (u + 9) = 0$

Step 2.2.3

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

$u - 4 = 0$

$u + 9 = 0$

Step 2.2.4

Set $u - 4$ equal to $0$ and solve for $u$ .

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Step 2.2.4.1

Set $u - 4$ equal to $0$ .

$u - 4 = 0$

Step 2.2.4.2

Add $4$ to both sides of the equation.

$u = 4$

Step 2.2.5

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

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Step 2.2.5.1

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

$u + 9 = 0$

Step 2.2.5.2

Subtract $9$ from both sides of the equation.

$u = - 9$

Step 2.2.6

The final solution is all the values that make $(u - 4) (u + 9) = 0$ true.

$u = 4, - 9$

Step 2.2.7

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

$x^{2} = 4$

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

Step 2.2.8

Solve the first equation for $x$ .

$x^{2} = 4$

Step 2.2.9

Solve the equation for $x$ .

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Step 2.2.9.1

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

$x = \pm \sqrt{4}$

Step 2.2.9.2

Simplify $\pm \sqrt{4}$ .

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Step 2.2.9.2.1

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

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

Step 2.2.9.2.2

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

$x = \pm 2$

Step 2.2.9.3

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

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Step 2.2.9.3.1

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

$x = 2$

Step 2.2.9.3.2

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

$x = - 2$

Step 2.2.9.3.3

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

$x = 2, - 2$

Step 2.2.10

Solve the second equation for $x$ .

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

Step 2.2.11

Solve the equation for $x$ .

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Step 2.2.11.1

Remove parentheses.

$x^{2} = - 9$

Step 2.2.11.2

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

$x = \pm \sqrt{- 9}$

Step 2.2.11.3

Simplify $\pm \sqrt{- 9}$ .

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Step 2.2.11.3.1

Rewrite $- 9$ as $- 1 (9)$ .

$x = \pm \sqrt{- 1 (9)}$

Step 2.2.11.3.2

Rewrite $\sqrt{- 1 (9)}$ as $\sqrt{- 1} \cdot \sqrt{9}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{9}$

Step 2.2.11.3.3

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

$x = \pm i \cdot \sqrt{9}$

Step 2.2.11.3.4

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

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

Step 2.2.11.3.5

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

$x = \pm i \cdot 3$

Step 2.2.11.3.6

Move $3$ to the left of $i$ .

$x = \pm 3 i$

Step 2.2.11.4

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

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Step 2.2.11.4.1

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

$x = 3 i$

Step 2.2.11.4.2

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

$x = - 3 i$

Step 2.2.11.4.3

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

$x = 3 i, - 3 i$

Step 2.2.12

The solution to $x^{4} + 5 x^{2} - 36 = 0$ is $x = 2, - 2, 3 i, - 3 i$ .

$x = 2, - 2, 3 i, - 3 i$

Step 3

Set $2 x^{2} + 9 x - 5$ equal to $0$ and solve for $x$ .

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Step 3.1

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

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

Step 3.2

Solve $2 x^{2} + 9 x - 5 = 0$ for $x$ .

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Step 3.2.1

Factor by grouping.

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Step 3.2.1.1

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 $a \cdot c = 2 \cdot - 5 = - 10$ and whose sum is $b = 9$ .

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Step 3.2.1.1.1

Factor $9$ out of $9 x$ .

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

Step 3.2.1.1.2

Rewrite $9$ as $- 1$ plus $10$

$2 x^{2} + (- 1 + 10) x - 5 = 0$

Step 3.2.1.1.3

Apply the distributive property.

$2 x^{2} - 1 x + 10 x - 5 = 0$

Step 3.2.1.2

Factor out the greatest common factor from each group.

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Step 3.2.1.2.1

Group the first two terms and the last two terms.

$(2 x^{2} - 1 x) + 10 x - 5 = 0$

Step 3.2.1.2.2

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

$x (2 x - 1) + 5 (2 x - 1) = 0$

Step 3.2.1.3

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$(2 x - 1) (x + 5) = 0$

Step 3.2.2

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

$2 x - 1 = 0$

$x + 5 = 0$

Step 3.2.3

Set $2 x - 1$ equal to $0$ and solve for $x$ .

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Step 3.2.3.1

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Step 3.2.3.2

Solve $2 x - 1 = 0$ for $x$ .

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Step 3.2.3.2.1

Add $1$ to both sides of the equation.

$2 x = 1$

Step 3.2.3.2.2

Divide each term in $2 x = 1$ by $2$ and simplify.

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Step 3.2.3.2.2.1

Divide each term in $2 x = 1$ by $2$ .

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

Step 3.2.3.2.2.2

Simplify the left side.

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Step 3.2.3.2.2.2.1

Cancel the common factor of $2$ .

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Step 3.2.3.2.2.2.1.1

Cancel the common factor.

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

Step 3.2.3.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 3.2.4

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

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Step 3.2.4.1

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

$x + 5 = 0$

Step 3.2.4.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 3.2.5

The final solution is all the values that make $(2 x - 1) (x + 5) = 0$ true.

$x = \frac{1}{2}, - 5$

Step 4

The final solution is all the values that make $(x^{4} + 5 x^{2} - 36) (2 x^{2} + 9 x - 5) = 0$ true.

$x = 2, - 2, 3 i, - 3 i, \frac{1}{2}, - 5$