Find the Roots (Zeros) x^4+49x^2

$x^{4} + 49 x^{2}$

Step 1

Set $x^{4} + 49 x^{2}$ equal to $0$ .

$x^{4} + 49 x^{2} = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

${(x^{2})}^{2} + 49 x^{2} = 0$

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$u^{2} + 49 u = 0$

Factor $u$ out of $u^{2} + 49 u$ .

Factor $u$ out of $u^{2}$ .

$u \cdot u + 49 u = 0$

Factor $u$ out of $49 u$ .

$u \cdot u + u \cdot 49 = 0$

Factor $u$ out of $u \cdot u + u \cdot 49$ .

$u (u + 49) = 0$

Replace all occurrences of $u$ with $x^{2}$ .

$x^{2} (x^{2} + 49) = 0$

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

$x^{2} = 0$

$x^{2} + 49 = 0$

Set $x^{2}$ equal to $0$ and solve for $x$ .

Set $x^{2}$ equal to $0$ .

$x^{2} = 0$

Solve $x^{2} = 0$ for $x$ .

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

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

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

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

Set $x^{2} + 49$ equal to $0$ and solve for $x$ .

Set $x^{2} + 49$ equal to $0$ .

$x^{2} + 49 = 0$

Solve $x^{2} + 49 = 0$ for $x$ .

Subtract $49$ from both sides of the equation.

$x^{2} = - 49$

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

$x = \pm \sqrt{- 49}$

Simplify $\pm \sqrt{- 49}$ .

Rewrite $- 49$ as $- 1 (49)$ .

$x = \pm \sqrt{- 1 (49)}$

Rewrite $\sqrt{- 1 (49)}$ as $\sqrt{- 1} \cdot \sqrt{49}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{49}$

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

$x = \pm i \cdot \sqrt{49}$

Rewrite $49$ as $7^{2}$ .

$x = \pm i \cdot \sqrt{7^{2}}$

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

$x = \pm i \cdot 7$

Move $7$ to the left of $i$ .

$x = \pm 7 i$

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.

$x = 7 i$

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

$x = - 7 i$

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

$x = 7 i, - 7 i$

The final solution is all the values that make $x^{2} (x^{2} + 49) = 0$ true.

$x = 0, 7 i, - 7 i$

Step 3

Find the Roots (Zeros) x^4+49x^2

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