Question

9x^2-6x+1=2 square root method

Answer 1

To solve the quadratic equation \( 9x^2 - 6x + 1 = 2 \) using the square root method, we first need to rearrange the equation into the standard form \( ax^2 + bx + c = 0 \).

### Step 1: Rearrange the equation
Start by moving all terms to one side of the equation:

\[
9x^2 - 6x + 1 - 2 = 0
\]

This simplifies to:

\[
9x^2 - 6x - 1 = 0
\]

### Step 2: Use the square root method
Next, we will use completing the square to solve this equation.

1. **Divide the entire equation by 9** (the coefficient of \(x^2\)):

\[
x^2 - \frac{2}{3}x - \frac{1}{9} = 0
\]

2. **Isolate the constant term**:

\[
x^2 - \frac{2}{3}x = \frac{1}{9}
\]

3. **Complete the square** on the left side. To do this, take half of the coefficient of \(x\), square it, and add it to both sides:

- Half of \(-\frac{2}{3}\) is \(-\frac{1}{3}\) and squaring it gives \(\left(-\frac{1}{3}\right)^2 = \frac{1}{9}\).

Adding \(\frac{1}{9}\) to both sides:

\[
x^2 - \frac{2}{3}x + \frac{1}{9} = \frac{1}{9} + \frac{1}{9}
\]

This simplifies to:

\[
\left(x - \frac{1}{3}\right)^2 = \frac{2}{9}
\]

### Step 3: Solve for \(x\) by taking the square root

Now, take the square root of both sides:

\[
x - \frac{1}{3} = \pm \sqrt{\frac{2}{9}}
\]

This leads to:

\[
x - \frac{1}{3} = \pm \frac{\sqrt{2}}{3}
\]

### Step 4: Solve for \(x\)

Now, solve for \(x\):

1. First case:

\[
x = \frac{1}{3} + \frac{\sqrt{2}}{3} = \frac{1 + \sqrt{2}}{3}
\]

2. Second case:

\[
x = \frac{1}{3} - \frac{\sqrt{2}}{3} = \frac{1 - \sqrt{2}}{3}
\]

### Final solutions
The solutions to the equation \( 9x^2 - 6x + 1 = 2 \) are:

\[
x = \frac{1 + \sqrt{2}}{3} \quad \text{and} \quad x = \frac{1 - \sqrt{2}}{3}
\]