Find the Roots (Zeros) (x^4+5x^2-36)(2x^2+9x-5)=0

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

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 $x$ .

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

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

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

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

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

$u = x^{2}$

Factor $u^{2} + 5 u - 36$ 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 $- 36$ and whose sum is $5$ .

$- 4, 9$

Write the factored form using these integers.

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

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

$u - 4 = 0$

$u + 9 = 0$

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

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

$u - 4 = 0$

Add $4$ to both sides of the equation.

$u = 4$

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

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

$u + 9 = 0$

Subtract $9$ from both sides of the equation.

$u = - 9$

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

$u = 4, - 9$

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

$x^{2} = 4$

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

Solve the first equation for $x$ .

$x^{2} = 4$

Solve the equation for $x$ .

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

$x = \pm \sqrt{4}$

Simplify $\pm \sqrt{4}$ .

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

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

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

$x = \pm 2$

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.

$x = 2$

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

$x = - 2$

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

$x = 2, - 2$

Solve the second equation for $x$ .

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

Solve the equation for $x$ .

Remove parentheses.

$x^{2} = - 9$

Take the specified 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 $- 1 (9)$ .

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

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

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

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

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

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

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

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

$x = \pm i \cdot 3$

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

$x = \pm 3 i$

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.

$x = 3 i$

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

$x = - 3 i$

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

$x = 3 i, - 3 i$

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 $x$ .

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

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

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

Factor by grouping.

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 $b = 9$ .

Factor $9$ out of $9 x$ .

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

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

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

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

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

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

$2 x - 1 = 0$

$x + 5 = 0$

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

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

$2 x - 1 = 0$

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

Add $1$ to both sides of the equation.

$2 x = 1$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

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.

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

Step 5

Find the Roots (Zeros) (x^4+5x^2-36)(2x^2+9x-5)=0

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