Solve by Factoring x^4-7x^2+6=0

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Algebra

$x^{4} - 7 x^{2} + 6 = 0$

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

${(x^{2})}^{2} - 7 x^{2} + 6 = 0$

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$u^{2} - 7 u + 6 = 0$

Factor $u^{2} - 7 u + 6$ using the AC method.

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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 $6$ and whose sum is $- 7$ .

$- 6, - 1$

Write the factored form using these integers.

$(u - 6) (u - 1) = 0$

Replace all occurrences of $u$ with $x^{2}$ .

$(x^{2} - 6) (x^{2} - 1) = 0$

Rewrite $1$ as $1^{2}$ .

$(x^{2} - 6) (x^{2} - 1^{2}) = 0$

Factor.

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Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$(x^{2} - 6) ((x + 1) (x - 1)) = 0$

Remove unnecessary parentheses.

$(x^{2} - 6) (x + 1) (x - 1) = 0$

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

$x^{2} - 6 = 0$

$x + 1 = 0$

$x - 1 = 0$

Set the first factor equal to $0$ and solve.

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Set the first factor equal to $0$ .

$x^{2} - 6 = 0$

Add $6$ to both sides of the equation.

$x^{2} = 6$

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

$x = \pm \sqrt{6}$

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

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First, use the positive value of the $\pm$ to find the first solution.

$x = \sqrt{6}$

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

$x = - \sqrt{6}$

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

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

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $(x^{2} - 6) (x + 1) (x - 1) = 0$ true.

$x = \sqrt{6}, - \sqrt{6}, - 1, 1$

The result can be shown in multiple forms.

Exact Form:

$x = \sqrt{6}, - \sqrt{6}, - 1, 1$

Decimal Form:

$x = 2.44948974 \dots, - 2.44948974 \dots, - 1, 1$

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