Find the Roots (Zeros) x^4+24x^2-25

$x^{4} + 24 x^{2} - 25$

Step 1

Set $x^{4} + 24 x^{2} - 25$ equal to $0$ .

$x^{4} + 24 x^{2} - 25 = 0$

Step 2

Solve for $x$ .

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

$u^{2} + 24 u - 25 = 0$

$u = x^{2}$

Factor $u^{2} + 24 u - 25$ 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 $- 25$ and whose sum is $24$ .

$- 1, 25$

Write the factored form using these integers.

$(u - 1) (u + 25) = 0$

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

$u - 1 = 0$

$u + 25 = 0$

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.

$u = 1$

Set $u + 25$ equal to $0$ and solve for $u$ .

Set $u + 25$ equal to $0$ .

$u + 25 = 0$

Subtract $25$ from both sides of the equation.

$u = - 25$

The final solution is all the values that make $(u - 1) (u + 25) = 0$ true.

$u = 1, - 25$

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

$x^{2} = 1$

${(x^{2})}^{1} = - 25$

Solve the first equation for $x$ .

$x^{2} = 1$

Solve the equation for $x$ .

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

$x = \pm \sqrt{1}$

Any root of $1$ is $1$ .

$x = \pm 1$

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.

$x = 1$

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

$x = - 1$

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

$x = 1, - 1$

Solve the second equation for $x$ .

${(x^{2})}^{1} = - 25$

Solve the equation for $x$ .

Remove parentheses.

$x^{2} = - 25$

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

$x = \pm \sqrt{- 25}$

Simplify $\pm \sqrt{- 25}$ .

Rewrite $- 25$ as $- 1 (25)$ .

$x = \pm \sqrt{- 1 (25)}$

Rewrite $\sqrt{- 1 (25)}$ as $\sqrt{- 1} \cdot \sqrt{25}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{25}$

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

$x = \pm i \cdot \sqrt{25}$

Rewrite $25$ as $5^{2}$ .

$x = \pm i \cdot \sqrt{5^{2}}$

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

$x = \pm i \cdot 5$

Move $5$ to the left of $i$ .

$x = \pm 5 i$

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.

$x = 5 i$

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

$x = - 5 i$

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

$x = 5 i, - 5 i$

The solution to $x^{4} + 24 x^{2} - 25 = 0$ is $x = 1, - 1, 5 i, - 5 i$ .

$x = 1, - 1, 5 i, - 5 i$

Step 3

Find the Roots (Zeros) x^4+24x^2-25

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