Question

7x^2+6x+2=0 by completing the square

Answer 1

To solve the quadratic equation \(7x^2 + 6x + 2 = 0\) by completing the square, follow these steps:

1. **Divide through by the coefficient of \(x^2\)**:
\[
x^2 + \frac{6}{7}x + \frac{2}{7} = 0
\]
(This simplifies our equation, and we have divided every term by \(7\).)

2. **Rearrange the equation**:
\[
x^2 + \frac{6}{7}x = -\frac{2}{7}
\]

3. **Complete the square**:
- First, take the coefficient of \(x\), which is \(\frac{6}{7}\), divide it by \(2\), and square it:
\[
\left(\frac{6}{14}\right)^2 = \left(\frac{3}{7}\right)^2 = \frac{9}{49}
\]

- Now add and subtract \(\frac{9}{49}\) on the left side:
\[
x^2 + \frac{6}{7}x + \frac{9}{49} - \frac{9}{49} = -\frac{2}{7}
\]
\[
x^2 + \frac{6}{7}x + \frac{9}{49} = -\frac{2}{7} + \frac{9}{49}
\]

4. **Simplify the right side**:
- To add the fractions, convert \(-\frac{2}{7}\) into a fraction with a denominator of \(49\):
\[
-\frac{2}{7} = -\frac{14}{49}
\]
Thus,
\[
-\frac{14}{49} + \frac{9}{49} = -\frac{5}{49}
\]

5. **Rewrite the left side as a square**:
\[
\left(x + \frac{3}{7}\right)^2 = -\frac{5}{49}
\]

6. **Take the square root of both sides**:
\[
x + \frac{3}{7} = \pm \sqrt{-\frac{5}{49}}
\]
\[
x + \frac{3}{7} = \pm \frac{\sqrt{5}i}{7} \quad (\text{because } \sqrt{-1} = i)
\]

7. **Solve for \(x\)**:
\[
x = -\frac{3}{7} \pm \frac{\sqrt{5}i}{7}
\]

8. **Express the solution**:
\[
x = -\frac{3 \pm \sqrt{5}i}{7}
\]

So the solutions to the equation \(7x^2 + 6x + 2 = 0\) are:
\[
x = -\frac{3 + \sqrt{5}i}{7}, \quad x = -\frac{3 - \sqrt{5}i}{7}
\]