Algebra Examples

5 i

$5 i$ ,

- 5 i

$- 5 i$

Step 1

x = 5 i

$x = 5 i$ and

x = - 5 i

$x = - 5 i$ are the two real distinct solutions for the quadratic equation, which means that

x - 5 i

$x - 5 i$ and

x - (- 5 i)

$x - (- 5 i)$ are the factors of the quadratic equation.

(x - 5 i) (x + 5 i) = 0

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

Step 2

Expand

(x - 5 i) (x + 5 i)

$(x - 5 i) (x + 5 i)$ using the FOIL Method.

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Step 2.1

Apply the distributive property.

x (x + 5 i) - 5 i (x + 5 i) = 0

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

Step 2.2

Apply the distributive property.

x \cdot x + x (5 i) - 5 i (x + 5 i) = 0

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

Step 2.3

Apply the distributive property.

x \cdot x + x (5 i) - 5 i x - 5 i (5 i) = 0

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

x \cdot x + x (5 i) - 5 i x - 5 i (5 i) = 0

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

Step 3

Simplify terms.

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Step 3.1

Combine the opposite terms in

x \cdot x + x (5 i) - 5 i x - 5 i (5 i)

$x \cdot x + x (5 i) - 5 i x - 5 i (5 i)$ .

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Step 3.1.1

Reorder the factors in the terms

x (5 i)

$x (5 i)$ and

- 5 i x

$- 5 i x$ .

x \cdot x + 5 i x - 5 i x - 5 i (5 i) = 0

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

Step 3.1.2

Subtract

5 i x

$5 i x$ from

5 i x

$5 i x$ .

x \cdot x + 0 - 5 i (5 i) = 0

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

Step 3.1.3

Add

x \cdot x

$x \cdot x$ and

0

$0$ .

x \cdot x - 5 i (5 i) = 0

$x \cdot x - 5 i (5 i) = 0$

x \cdot x - 5 i (5 i) = 0

$x \cdot x - 5 i (5 i) = 0$

Step 3.2

Simplify each term.

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Step 3.2.1

Multiply

x

$x$ by

x

$x$ .

x^{2} - 5 i (5 i) = 0

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

Step 3.2.2

Multiply

- 5 i (5 i)

$- 5 i (5 i)$ .

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Step 3.2.2.1

Multiply

5

$5$ by

- 5

$- 5$ .

x^{2} - 25 i i = 0

$x^{2} - 25 i i = 0$

Step 3.2.2.2

Raise

i

$i$ to the power of

1

$1$ .

x^{2} - 25 (i i) = 0

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

Step 3.2.2.3

Raise

i

$i$ to the power of

1

$1$ .

x^{2} - 25 (i i) = 0

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

Step 3.2.2.4

Use the power rule

a^{m} a^{n} = a^{m + n}

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

x^{2} - 25 i^{1 + 1} = 0

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

Step 3.2.2.5

Add

1

$1$ and

1

$1$ .

x^{2} - 25 i^{2} = 0

$x^{2} - 25 i^{2} = 0$

x^{2} - 25 i^{2} = 0

$x^{2} - 25 i^{2} = 0$

Step 3.2.3

Rewrite

i^{2}

$i^{2}$ as

- 1

$- 1$ .

x^{2} - 25 \cdot - 1 = 0

$x^{2} - 25 \cdot - 1 = 0$

Step 3.2.4

Multiply

- 25

$- 25$ by

- 1

$- 1$ .

x^{2} + 25 = 0

$x^{2} + 25 = 0$

x^{2} + 25 = 0

$x^{2} + 25 = 0$

x^{2} + 25 = 0

$x^{2} + 25 = 0$

Step 4

The standard quadratic equation using the given set of solutions

{5 i, - 5 i}

${5 i, - 5 i}$ is

y = x^{2} + 25

$y = x^{2} + 25$ .

y = x^{2} + 25

$y = x^{2} + 25$

Step 5