Find the Roots (Zeros) x^4+10x^2+25=0

$x^{4} + 10 x^{2} + 25 = 0$

Step 1

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

$u^{2} + 10 u + 25 = 0$

$u = x^{2}$

Step 2

Factor using the perfect square rule.

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

$u^{2} + 10 u + 5^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$10 u = 2 \cdot u \cdot 5$

Rewrite the polynomial.

$u^{2} + 2 \cdot u \cdot 5 + 5^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = u$ and $b = 5$ .

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

Step 3

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

$u + 5 = 0$

Step 4

Subtract $5$ from both sides of the equation.

$u = - 5$

Step 5

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

$x^{2} = - 5$

Step 6

Solve the equation for $x$ .

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

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

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

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

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

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

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

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

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

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.

$x = i \sqrt{5}$

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

$x = - i \sqrt{5}$

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

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

Step 7