Algebra Examples

$- 5 + 2 i$ , $- 5 i$

Step 1

$x = - 5 + 2 i$ and $x = - 5 i$ are the two real distinct solutions for the quadratic equation, which means that $x - - 5 + 2 i$ and $x - (- 5 i)$ are the factors of the quadratic equation.

$(x + 5 + 2 i) (x + 5 i) = 0$

Step 2

Expand $(x + 5 + 2 i) (x + 5 i)$ by multiplying each term in the first expression by each term in the second expression.

$x \cdot x + x (5 i) + 5 x + 5 (5 i) + 2 i x + 2 i (5 i) = 0$

Step 3

Simplify each term.

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Step 3.1

Multiply $x$ by $x$ .

$x^{2} + x (5 i) + 5 x + 5 (5 i) + 2 i x + 2 i (5 i) = 0$

Step 3.2

Move $5$ to the left of $x$ .

$x^{2} + 5 \cdot (x i) + 5 x + 5 (5 i) + 2 i x + 2 i (5 i) = 0$

Step 3.3

Multiply $5$ by $5$ .

$x^{2} + 5 x i + 5 x + 25 i + 2 i x + 2 i (5 i) = 0$

Step 3.4

Multiply $2 i (5 i)$ .

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Step 3.4.1

Multiply $5$ by $2$ .

$x^{2} + 5 x i + 5 x + 25 i + 2 i x + 10 i i = 0$

Step 3.4.2

Raise $i$ to the power of $1$ .

$x^{2} + 5 x i + 5 x + 25 i + 2 i x + 10 (i i) = 0$

Step 3.4.3

Raise $i$ to the power of $1$ .

$x^{2} + 5 x i + 5 x + 25 i + 2 i x + 10 (i i) = 0$

Step 3.4.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x^{2} + 5 x i + 5 x + 25 i + 2 i x + 10 i^{1 + 1} = 0$

Step 3.4.5

Add $1$ and $1$ .

$x^{2} + 5 x i + 5 x + 25 i + 2 i x + 10 i^{2} = 0$

Step 3.5

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

$x^{2} + 5 x i + 5 x + 25 i + 2 i x + 10 \cdot - 1 = 0$

Step 3.6

Multiply $10$ by $- 1$ .

$x^{2} + 5 x i + 5 x + 25 i + 2 i x - 10 = 0$

Step 4

Add $5 x i$ and $2 i x$ .

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Step 4.1

Move $i$ .

$x^{2} + 5 x i + 2 x i + 5 x + 25 i - 10 = 0$

Step 4.2

Add $5 x i$ and $2 x i$ .

$x^{2} + 7 x i + 5 x + 25 i - 10 = 0$

Step 5

The standard quadratic equation using the given set of solutions ${- 5 + 2 i, - 5 i}$ is $y = x^{2} + 7 x i + 5 x + 25 i - 10$ .

$y = x^{2} + 7 x i + 5 x + 25 i - 10$

Step 6