Find the Quadratic Equation 4-3i , 4+3i

$4 - 3 i$ , $4 + 3 i$

Step 1

$x = 4 - 3 i$ and $x = 4 + 3 i$ are the two real distinct solutions for the quadratic equation, which means that $x - 4 - 3 i$ and $x - 4 + 3 i$ are the factors of the quadratic equation.

$(x - 4 - 3 i) (x - 4 + 3 i) = 0$

Step 2

Expand $(x - 4 - 3 i) (x - 4 + 3 i)$ by multiplying each term in the first expression by each term in the second expression.

$x \cdot x + x \cdot - 4 + x (3 i) - 4 x - 4 \cdot - 4 - 4 (3 i) - 3 i x - 3 i \cdot - 4 - 3 i (3 i) = 0$

Step 3

Simplify terms.

Combine the opposite terms in $x \cdot x + x \cdot - 4 + x (3 i) - 4 x - 4 \cdot - 4 - 4 (3 i) - 3 i x - 3 i \cdot - 4 - 3 i (3 i)$ .

Reorder the factors in the terms $x (3 i)$ and $- 3 i x$ .

$x \cdot x + x \cdot - 4 + 3 i x - 4 x - 4 \cdot - 4 - 4 (3 i) - 3 i x - 3 i \cdot - 4 - 3 i (3 i) = 0$

Subtract $3 i x$ from $3 i x$ .

$x \cdot x + x \cdot - 4 + 0 - 4 x - 4 \cdot - 4 - 4 (3 i) - 3 i \cdot - 4 - 3 i (3 i) = 0$

Add $x \cdot x + x \cdot - 4$ and $0$ .

$x \cdot x + x \cdot - 4 - 4 x - 4 \cdot - 4 - 4 (3 i) - 3 i \cdot - 4 - 3 i (3 i) = 0$

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 4 - 4 x - 4 \cdot - 4 - 4 (3 i) - 3 i \cdot - 4 - 3 i (3 i) = 0$

Move $- 4$ to the left of $x$ .

$x^{2} - 4 \cdot x - 4 x - 4 \cdot - 4 - 4 (3 i) - 3 i \cdot - 4 - 3 i (3 i) = 0$

Multiply $- 4$ by $- 4$ .

$x^{2} - 4 x - 4 x + 16 - 4 (3 i) - 3 i \cdot - 4 - 3 i (3 i) = 0$

Multiply $3$ by $- 4$ .

$x^{2} - 4 x - 4 x + 16 - 12 i - 3 i \cdot - 4 - 3 i (3 i) = 0$

Multiply $- 4$ by $- 3$ .

$x^{2} - 4 x - 4 x + 16 - 12 i + 12 i - 3 i (3 i) = 0$

Multiply $- 3 i (3 i)$ .

Multiply $3$ by $- 3$ .

$x^{2} - 4 x - 4 x + 16 - 12 i + 12 i - 9 i i = 0$

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

$x^{2} - 4 x - 4 x + 16 - 12 i + 12 i - 9 (i i) = 0$

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

$x^{2} - 4 x - 4 x + 16 - 12 i + 12 i - 9 (i i) = 0$

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

$x^{2} - 4 x - 4 x + 16 - 12 i + 12 i - 9 i^{1 + 1} = 0$

Add $1$ and $1$ .

$x^{2} - 4 x - 4 x + 16 - 12 i + 12 i - 9 i^{2} = 0$

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

$x^{2} - 4 x - 4 x + 16 - 12 i + 12 i - 9 \cdot - 1 = 0$

Multiply $- 9$ by $- 1$ .

$x^{2} - 4 x - 4 x + 16 - 12 i + 12 i + 9 = 0$

Simplify by adding terms.

Combine the opposite terms in $x^{2} - 4 x - 4 x + 16 - 12 i + 12 i + 9$ .

Add $- 12 i$ and $12 i$ .

$x^{2} - 4 x - 4 x + 16 + 0 + 9 = 0$

Add $x^{2} - 4 x - 4 x + 16$ and $0$ .

$x^{2} - 4 x - 4 x + 16 + 9 = 0$

Subtract $4 x$ from $- 4 x$ .

$x^{2} - 8 x + 16 + 9 = 0$

Add $16$ and $9$ .

$x^{2} - 8 x + 25 = 0$

Step 4

The standard quadratic equation using the given set of solutions ${4 - 3 i, 4 + 3 i}$ is $y = x^{2} - 8 x + 25$ .

$y = x^{2} - 8 x + 25$

Step 5

Find the Quadratic Equation 4-3i , 4+3i

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