Solve for a |a^2-3a+6|=|a-3|

$| a^{2} - 3 a + 6 | = | a - 3 |$

Step 1

Rewrite the equation as $| a - 3 | = | a^{2} - 3 a + 6 |$ .

$| a - 3 | = | a^{2} - 3 a + 6 |$

Step 2

Rewrite the absolute value equation as four equations without absolute value bars.

$a - 3 = a^{2} - 3 a + 6$

$a - 3 = - (a^{2} - 3 a + 6)$

$- (a - 3) = a^{2} - 3 a + 6$

$- (a - 3) = - (a^{2} - 3 a + 6)$

Step 3

After simplifying, there are only two unique equations to be solved.

$a - 3 = a^{2} - 3 a + 6$

$a - 3 = - (a^{2} - 3 a + 6)$

Step 4

Solve $a - 3 = a^{2} - 3 a + 6$ for $a$ .

Since $a$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$a^{2} - 3 a + 6 = a - 3$

Move all terms containing $a$ to the left side of the equation.

Subtract $a$ from both sides of the equation.

$a^{2} - 3 a + 6 - a = - 3$

Subtract $a$ from $- 3 a$ .

$a^{2} - 4 a + 6 = - 3$

Move all terms to the left side of the equation and simplify.

Add $3$ to both sides of the equation.

$a^{2} - 4 a + 6 + 3 = 0$

Add $6$ and $3$ .

$a^{2} - 4 a + 9 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 4$ , and $c = 9$ into the quadratic formula and solve for $a$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (1 \cdot 9)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $- 4$ to the power of $2$ .

$a = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot 9}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 9$ .

Multiply $- 4$ by $1$ .

$a = \frac{4 \pm \sqrt{16 - 4 \cdot 9}}{2 \cdot 1}$

Multiply $- 4$ by $9$ .

$a = \frac{4 \pm \sqrt{16 - 36}}{2 \cdot 1}$

Subtract $36$ from $16$ .

$a = \frac{4 \pm \sqrt{- 20}}{2 \cdot 1}$

Rewrite $- 20$ as $- 1 (20)$ .

$a = \frac{4 \pm \sqrt{- 1 \cdot 20}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (20)}$ as $\sqrt{- 1} \cdot \sqrt{20}$ .

$a = \frac{4 \pm \sqrt{- 1} \cdot \sqrt{20}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$a = \frac{4 \pm i \cdot \sqrt{20}}{2 \cdot 1}$

Rewrite $20$ as $2^{2} \cdot 5$ .

Factor $4$ out of $20$ .

$a = \frac{4 \pm i \cdot \sqrt{4 (5)}}{2 \cdot 1}$

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

$a = \frac{4 \pm i \cdot \sqrt{2^{2} \cdot 5}}{2 \cdot 1}$

Pull terms out from under the radical.

$a = \frac{4 \pm i \cdot (2 \sqrt{5})}{2 \cdot 1}$

Move $2$ to the left of $i$ .

$a = \frac{4 \pm 2 i \sqrt{5}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$a = \frac{4 \pm 2 i \sqrt{5}}{2}$

Simplify $\frac{4 \pm 2 i \sqrt{5}}{2}$ .

$a = 2 \pm i \sqrt{5}$

The final answer is the combination of both solutions.

$a = 2 + i \sqrt{5}, 2 - i \sqrt{5}$

Step 5

Solve $a - 3 = - (a^{2} - 3 a + 6)$ for $a$ .

Since $a$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$- (a^{2} - 3 a + 6) = a - 3$

Simplify $- (a^{2} - 3 a + 6)$ .

Rewrite.

$0 + 0 - (a^{2} - 3 a + 6) = a - 3$

Simplify by adding zeros.

$- (a^{2} - 3 a + 6) = a - 3$

Apply the distributive property.

$- a^{2} - (- 3 a) - 1 \cdot 6 = a - 3$

Simplify.

Multiply $- 3$ by $- 1$ .

$- a^{2} + 3 a - 1 \cdot 6 = a - 3$

Multiply $- 1$ by $6$ .

$- a^{2} + 3 a - 6 = a - 3$

Move all terms containing $a$ to the left side of the equation.

Subtract $a$ from both sides of the equation.

$- a^{2} + 3 a - 6 - a = - 3$

Subtract $a$ from $3 a$ .

$- a^{2} + 2 a - 6 = - 3$

Move all terms to the left side of the equation and simplify.

Add $3$ to both sides of the equation.

$- a^{2} + 2 a - 6 + 3 = 0$

Add $- 6$ and $3$ .

$- a^{2} + 2 a - 3 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = - 1$ , $b = 2$ , and $c = - 3$ into the quadratic formula and solve for $a$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (- 1 \cdot - 3)}}{2 \cdot - 1}$

Simplify.

Simplify the numerator.

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

$a = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 1 \cdot - 3}}{2 \cdot - 1}$

Multiply $- 4 \cdot - 1 \cdot - 3$ .

Multiply $- 4$ by $- 1$ .

$a = \frac{- 2 \pm \sqrt{4 + 4 \cdot - 3}}{2 \cdot - 1}$

Multiply $4$ by $- 3$ .

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

Subtract $12$ from $4$ .

$a = \frac{- 2 \pm \sqrt{- 8}}{2 \cdot - 1}$

Rewrite $- 8$ as $- 1 (8)$ .

$a = \frac{- 2 \pm \sqrt{- 1 \cdot 8}}{2 \cdot - 1}$

Rewrite $\sqrt{- 1 (8)}$ as $\sqrt{- 1} \cdot \sqrt{8}$ .

$a = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{8}}{2 \cdot - 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$a = \frac{- 2 \pm i \cdot \sqrt{8}}{2 \cdot - 1}$

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

$a = \frac{- 2 \pm i \cdot \sqrt{4 (2)}}{2 \cdot - 1}$

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

$a = \frac{- 2 \pm i \cdot \sqrt{2^{2} \cdot 2}}{2 \cdot - 1}$

Pull terms out from under the radical.

$a = \frac{- 2 \pm i \cdot (2 \sqrt{2})}{2 \cdot - 1}$

Move $2$ to the left of $i$ .

$a = \frac{- 2 \pm 2 i \sqrt{2}}{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$a = \frac{- 2 \pm 2 i \sqrt{2}}{- 2}$

Simplify $\frac{- 2 \pm 2 i \sqrt{2}}{- 2}$ .

$a = 1 \pm i \sqrt{2}$

The final answer is the combination of both solutions.

$a = 1 + i \sqrt{2}, 1 - i \sqrt{2}$

Step 6

List all of the solutions.

$a =$

Solve for a |a^2-3a+6|=|a-3|

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