Identify the Zeros and Their Multiplicities f(x)=x^4-27x^2+50

$f (x) = x^{4} - 27 x^{2} + 50$

Step 1

Set $x^{4} - 27 x^{2} + 50$ equal to $0$ .

$x^{4} - 27 x^{2} + 50 = 0$

Step 2

Solve for $x$ .

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

$u^{2} - 27 u + 50 = 0$

$u = x^{2}$

Factor $u^{2} - 27 u + 50$ 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 $50$ and whose sum is $- 27$ .

$- 25, - 2$

Write the factored form using these integers.

$(u - 25) (u - 2) = 0$

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

$u - 25 = 0$

$u - 2 = 0$

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

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

$u - 25 = 0$

Add $25$ to both sides of the equation.

$u = 25$

Set $u - 2$ equal to $0$ and solve for $u$ .

Set $u - 2$ equal to $0$ .

$u - 2 = 0$

Add $2$ to both sides of the equation.

$u = 2$

The final solution is all the values that make $(u - 25) (u - 2) = 0$ true. The multiplicity of a root is the number of times the root appears.

$u = 25$ (Multiplicity of $1$ )

$u = 2$ (Multiplicity of $1$ )

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

$x^{2} = 25$

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

Solve the first equation for $x$ .

$x^{2} = 25$

Solve the equation for $x$ .

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

$x = \pm \sqrt{25}$

Simplify $\pm \sqrt{25}$ .

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

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

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

$x = \pm 5$

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.

$x = 5$

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

$x = - 5$

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

$x = 5, - 5$

The multiplicity of a root is the number of times the root appears. For example, a factor of ${(x + 5)}^{3}$ would have a root at $x = - 5$ with multiplicity of $3$ .

$x = 5$ (Multiplicity of $1$ )

$x = - 5$ (Multiplicity of $1$ )

$x = 5$ (Multiplicity of $1$ )

$x = - 5$ (Multiplicity of $1$ )

Solve the second equation for $x$ .

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

Solve the equation for $x$ .

Remove parentheses.

$x^{2} = 2$

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

$x = \pm \sqrt{2}$

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.

$x = \sqrt{2}$

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

$x = - \sqrt{2}$

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

$x = \sqrt{2}, - \sqrt{2}$

The multiplicity of a root is the number of times the root appears. For example, a factor of ${(x + 5)}^{3}$ would have a root at $x = - 5$ with multiplicity of $3$ .

$x = \sqrt{2}$ (Multiplicity of $1$ )

$x = - \sqrt{2}$ (Multiplicity of $1$ )

$x = \sqrt{2}$ (Multiplicity of $1$ )

$x = - \sqrt{2}$ (Multiplicity of $1$ )

The solution to $x^{4} - 27 x^{2} + 50 = 0$ is $x = 5, - 5, \sqrt{2}, - \sqrt{2}$ .

$x = 5, - 5, \sqrt{2}, - \sqrt{2}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = 5, - 5, \sqrt{2}, - \sqrt{2}$

Decimal Form:

$x = 5, - 5, 1.41421356 \dots, - 1.41421356 \dots$

Step 4

Identify the Zeros and Their Multiplicities f(x)=x^4-27x^2+50

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