Solve for a a^4-13a^2+42=0

$a^{4} - 13 a^{2} + 42 = 0$

Step 1

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

$u^{2} - 13 u + 42 = 0$

$u = a^{2}$

Step 2

Factor $u^{2} - 13 u + 42$ 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 $42$ and whose sum is $- 13$ .

$- 7, - 6$

Write the factored form using these integers.

$(u - 7) (u - 6) = 0$

Step 3

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

$u - 7 = 0$

$u - 6 = 0$

Step 4

Set $u - 7$ equal to $0$ and solve for $u$ .

Set $u - 7$ equal to $0$ .

$u - 7 = 0$

Add $7$ to both sides of the equation.

$u = 7$

Step 5

Set $u - 6$ equal to $0$ and solve for $u$ .

Set $u - 6$ equal to $0$ .

$u - 6 = 0$

Add $6$ to both sides of the equation.

$u = 6$

Step 6

The final solution is all the values that make $(u - 7) (u - 6) = 0$ true.

$u = 7, 6$

Step 7

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

$a^{2} = 7$

${(a^{2})}^{1} = 6$

Step 8

Solve the first equation for $a$ .

$a^{2} = 7$

Step 9

Solve the equation for $a$ .

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

$a = \pm \sqrt{7}$

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.

$a = \sqrt{7}$

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

$a = - \sqrt{7}$

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

$a = \sqrt{7}, - \sqrt{7}$

Step 10

Solve the second equation for $a$ .

${(a^{2})}^{1} = 6$

Step 11

Solve the equation for $a$ .

Remove parentheses.

$a^{2} = 6$

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

$a = \pm \sqrt{6}$

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.

$a = \sqrt{6}$

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

$a = - \sqrt{6}$

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

$a = \sqrt{6}, - \sqrt{6}$

Step 12

The solution to $a^{4} - 13 a^{2} + 42 = 0$ is $a = \sqrt{7}, - \sqrt{7}, \sqrt{6}, - \sqrt{6}$ .

$a = \sqrt{7}, - \sqrt{7}, \sqrt{6}, - \sqrt{6}$

Step 13

The result can be shown in multiple forms.

Exact Form:

$a = \sqrt{7}, - \sqrt{7}, \sqrt{6}, - \sqrt{6}$

Decimal Form:

$a = 2.64575131 \dots, - 2.64575131 \dots, 2.44948974 \dots, - 2.44948974 \dots$

Solve for a a^4-13a^2+42=0

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