Precalculus Examples

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

Step 1

Rewrite the equation as $x^{2} - 2 x - 2 = j (- 1)$ .

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

Step 2

Simplify $j (- 1)$ .

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Step 2.1

Move $- 1$ to the left of $j$ .

$x^{2} - 2 x - 2 = - 1 j$

Step 2.2

Rewrite $- 1 j$ as $- j$ .

$x^{2} - 2 x - 2 = - j$

Step 3

Add $j$ to both sides of the equation.

$x^{2} - 2 x - 2 + j = 0$

Step 4

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 5

Substitute the values $a = 1$ , $b = - 2$ , and $c = - 2 + j$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (1 \cdot (- 2 + j))}}{2 \cdot 1}$

Step 6

Simplify.

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Step 6.1

Simplify the numerator.

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Step 6.1.1

Raise $- 2$ to the power of $2$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot (- 2 + j)}}{2 \cdot 1}$

Step 6.1.2

Multiply $- 4$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot (- 2 + j)}}{2 \cdot 1}$

Step 6.1.3

Apply the distributive property.

$x = \frac{2 \pm \sqrt{4 - 4 \cdot - 2 - 4 j}}{2 \cdot 1}$

Step 6.1.4

Multiply $- 4$ by $- 2$ .

$x = \frac{2 \pm \sqrt{4 + 8 - 4 j}}{2 \cdot 1}$

Step 6.1.5

Add $4$ and $8$ .

$x = \frac{2 \pm \sqrt{12 - 4 j}}{2 \cdot 1}$

Step 6.1.6

Factor $4$ out of $12 - 4 j$ .

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Step 6.1.6.1

Factor $4$ out of $12$ .

$x = \frac{2 \pm \sqrt{4 (3) - 4 j}}{2 \cdot 1}$

Step 6.1.6.2

Factor $4$ out of $- 4 j$ .

$x = \frac{2 \pm \sqrt{4 (3) + 4 (- j)}}{2 \cdot 1}$

Step 6.1.6.3

Factor $4$ out of $4 (3) + 4 (- j)$ .

$x = \frac{2 \pm \sqrt{4 (3 - j)}}{2 \cdot 1}$

Step 6.1.7

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

$x = \frac{2 \pm \sqrt{2^{2} (3 - j)}}{2 \cdot 1}$

Step 6.1.8

Pull terms out from under the radical.

$x = \frac{2 \pm 2 \sqrt{3 - j}}{2 \cdot 1}$

Step 6.2

Multiply $2$ by $1$ .

$x = \frac{2 \pm 2 \sqrt{3 - j}}{2}$

Step 6.3

Simplify $\frac{2 \pm 2 \sqrt{3 - j}}{2}$ .

$x = 1 \pm \sqrt{3 - j}$

Step 7

The final answer is the combination of both solutions.

$x = 1 + \sqrt{3 - j}$

$x = 1 - \sqrt{3 - j}$