Solve for p p^4+40=14p^2

$p^{4} + 40 = 14 p^{2}$

Subtract $14 p^{2}$ from both sides of the equation.

$p^{4} + 40 - 14 p^{2} = 0$

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

$u^{2} - 14 u + 40 = 0$

$u = p^{2}$

Factor $u^{2} - 14 u + 40$ 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 $40$ and whose sum is $- 14$ .

$- 10, - 4$

Write the factored form using these integers.

$(u - 10) (u - 4) = 0$

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

$u - 10 = 0$

$u - 4 = 0$

Set $u - 10$ equal to $0$ and solve for $u$ .

Set $u - 10$ equal to $0$ .

$u - 10 = 0$

Add $10$ to both sides of the equation.

$u = 10$

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.

$u = 4$

The final solution is all the values that make $(u - 10) (u - 4) = 0$ true.

$u = 10, 4$

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

$p^{2} = 10$

${(p^{2})}^{1} = 4$

Solve the first equation for $p$ .

$p^{2} = 10$

Solve the equation for $p$ .

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

$p = \pm \sqrt{10}$

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.

$p = \sqrt{10}$

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

$p = - \sqrt{10}$

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

$p = \sqrt{10}, - \sqrt{10}$

Solve the second equation for $p$ .

${(p^{2})}^{1} = 4$

Solve the equation for $p$ .

Remove parentheses.

$p^{2} = 4$

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

$p = \pm \sqrt{4}$

Simplify $\pm \sqrt{4}$ .

Rewrite $4$ as $2^{2}$ .

$p = \pm \sqrt{2^{2}}$

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

$p = \pm 2$

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.

$p = 2$

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

$p = - 2$

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

$p = 2, - 2$

The solution to $p^{4} - 14 p^{2} + 40 = 0$ is $p = \sqrt{10}, - \sqrt{10}, 2, - 2$ .

$p = \sqrt{10}, - \sqrt{10}, 2, - 2$

The result can be shown in multiple forms.

Exact Form:

$p = \sqrt{10}, - \sqrt{10}, 2, - 2$

Decimal Form:

$p = 3.16227766 \dots, - 3.16227766 \dots, 2, - 2$

Solve for p p^4+40=14p^2

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