Question

x^2+3x+6=0 by completing the square

Answer 1

To solve the quadratic equation \(x^2 + 3x + 6 = 0\) by completing the square, follow these steps:

1. **Move the constant term to the right side**:
\[
x^2 + 3x = -6
\]

2. **Complete the square**:
Take the coefficient of \(x\) (which is 3), divide it by 2, and square it:
\[
\left(\frac{3}{2}\right)^2 = \frac{9}{4}
\]

3. **Add \(\frac{9}{4}\) to both sides**:
\[
x^2 + 3x + \frac{9}{4} = -6 + \frac{9}{4}
\]

To simplify the right side, convert \(-6\) to a fraction with a denominator of 4:
\[
-6 = -\frac{24}{4}
\]
Therefore,
\[
-6 + \frac{9}{4} = -\frac{24}{4} + \frac{9}{4} = -\frac{15}{4}
\]

4. **Rewrite the left side as a square**:
\[
\left(x + \frac{3}{2}\right)^2 = -\frac{15}{4}
\]

5. **Take the square root of both sides**:
\[
x + \frac{3}{2} = \pm \sqrt{-\frac{15}{4}}
\]
This can be rewritten as:
\[
x + \frac{3}{2} = \pm \frac{\sqrt{15}}{2} i
\]

6. **Isolate \(x\)**:
\[
x = -\frac{3}{2} \pm \frac{\sqrt{15}}{2} i
\]

So, the solutions to the equation \(x^2 + 3x + 6 = 0\) are:
\[
x = -\frac{3}{2} + \frac{\sqrt{15}}{2} i \quad \text{and} \quad x = -\frac{3}{2} - \frac{\sqrt{15}}{2} i
\]