Find the Roots (Zeros) f(x)=x^4-5x^2+4

$f (x) = x^{4} - 5 x^{2} + 4$

Step 1

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

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

Step 2

Solve for $x$ .

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

$u^{2} - 5 u + 4 = 0$

$u = x^{2}$

Factor $u^{2} - 5 u + 4$ 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 $4$ and whose sum is $- 5$ .

$- 4, - 1$

Write the factored form using these integers.

$(u - 4) (u - 1) = 0$

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

$u - 4 = 0$

$u - 1 = 0$

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 $u - 1$ equal to $0$ and solve for $u$ .

Set $u - 1$ equal to $0$ .

$u - 1 = 0$

Add $1$ to both sides of the equation.

$u = 1$

The final solution is all the values that make $(u - 4) (u - 1) = 0$ true.

$u = 4, 1$

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

$x^{2} = 4$

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

Solve the first equation for $x$ .

$x^{2} = 4$

Solve the equation for $x$ .

Take the square 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 $x$ .

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

Solve the equation for $x$ .

Remove parentheses.

$x^{2} = 1$

Take the square 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.

$x = 1, - 1$

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

$x = 2, - 2, 1, - 1$

Step 3

Find the Roots (Zeros) f(x)=x^4-5x^2+4

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