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.
- Divide the entire equation by 9 (the coefficient of \(x^2\)):
\[ x^2 - \frac{2}{3}x - \frac{1}{9} = 0 \]
- Isolate the constant term:
\[ x^2 - \frac{2}{3}x = \frac{1}{9} \]
- 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 \(x\):
- First case:
\[ x = \frac{1}{3} + \frac{\sqrt{2}}{3} = \frac{1 + \sqrt{2}}{3} \]
- 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:
\[ x = \frac{1 + \sqrt{2}}{3} \quad \text{and} \quad x = \frac{1 - \sqrt{2}}{3} \]
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