Find All Complex Solutions b^2-4ac=0

$b^{2} - 4 a c = 0$

Rewrite $b^{2} - 4 a c$ as a difference of squares.

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = b$ and $b = 2 \sqrt{a c}$ .

$(b + 2 \sqrt{a c}) (b - (2 \sqrt{a c}))$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$b + 2 \sqrt{a c} = 0$

$b - 1 \cdot (2 \sqrt{a c}) = 0$

Set $b + 2 \sqrt{a c}$ equal to $0$ and solve for $b$ .

Set $b + 2 \sqrt{a c}$ equal to $0$ .

$b + 2 \sqrt{a c} = 0$

Subtract $2 \sqrt{a c}$ from both sides of the equation.

$b = - 2 \sqrt{a c}$

Set $b - 1 \cdot (2 \sqrt{a c})$ equal to $0$ and solve for $b$ .

Set $b - 1 \cdot (2 \sqrt{a c})$ equal to $0$ .

$b - 1 \cdot (2 \sqrt{a c}) = 0$

Solve $b - 1 \cdot (2 \sqrt{a c}) = 0$ for $b$ .

Multiply $2$ by $- 1$ .

$b - 2 \sqrt{a c} = 0$

Add $2 \sqrt{a c}$ to both sides of the equation.

$b = 2 \sqrt{a c}$

The final solution is all the values that make $(b + 2 \sqrt{a c}) (b - 1 \cdot (2 \sqrt{a c})) = 0$ true.

$b = - 2 \sqrt{a c}$

$b = 2 \sqrt{a c}$

Find All Complex Solutions b^2-4ac=0

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