Question

factor and solve. remember to check for a gcf first, and there should be 4 solutions for each.

1. ( x^4 - 6x^2 - 27 = 0 )
2. ( x^4 - 9x^2 + 14 = 0 )

Explanation:

For problem 3: $x^4 - 6x^2 - 27 = 0$

Step1: Substitute $u=x^2$

Let $u = x^2$, so the equation becomes:
$$u^2 - 6u - 27 = 0$$

Step2: Factor quadratic in $u$

Find two numbers that multiply to $-27$ and add to $-6$: $-9$ and $3$.
$$(u - 9)(u + 3) = 0$$

Step3: Solve for $u$

Set each factor equal to 0:
$u - 9 = 0 \implies u = 9$
$u + 3 = 0 \implies u = -3$

Step4: Substitute back $u=x^2$

For $u=9$: $x^2 = 9 \implies x = \pm 3$
For $u=-3$: $x^2 = -3 \implies x = \pm i\sqrt{3}$

Step1: Substitute $u=x^2$

Let $u = x^2$, so the equation becomes:
$$u^2 - 9u + 14 = 0$$

Step2: Factor quadratic in $u$

Find two numbers that multiply to $14$ and add to $-9$: $-7$ and $-2$.
$$(u - 7)(u - 2) = 0$$

Step3: Solve for $u$

Set each factor equal to 0:
$u - 7 = 0 \implies u = 7$
$u - 2 = 0 \implies u = 2$

Step4: Substitute back $u=x^2$

For $u=7$: $x^2 = 7 \implies x = \pm \sqrt{7}$
For $u=2$: $x^2 = 2 \implies x = \pm \sqrt{2}$

Answer:

$x = 3,\ x = -3,\ x = i\sqrt{3},\ x = -i\sqrt{3}$

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For problem 4: $x^4 - 9x^2 + 14 = 0$