To solve this problem, we need to find the quadratic equation given the roots \(\frac{1}{3+\sqrt{2}}\) and \(\frac{1}{3-\sqrt{2}}\). Let's begin by simplifying these roots:
- Rationalize the roots:
\(x_1 = \frac{1}{3+\sqrt{2}} = \frac{1}{3+\sqrt{2}} \times \frac{3-\sqrt{2}}{3-\sqrt{2}} = \frac{3-\sqrt{2}}{9-2} = \frac{3-\sqrt{2}}{7}\)
\(x_2 = \frac{1}{3-\sqrt{2}} = \frac{1}{3-\sqrt{2}} \times \frac{3+\sqrt{2}}{3+\sqrt{2}} = \frac{3+\sqrt{2}}{9-2} = \frac{3+\sqrt{2}}{7}\) - Sum and product of roots:
- Sum of roots \((s)\) is given by:
\(s = x_1 + x_2 = \frac{3-\sqrt{2}}{7} + \frac{3+\sqrt{2}}{7} = \frac{6}{7}\) - Product of roots \((p)\) is given by:
\(p = x_1 \cdot x_2 = \left(\frac{3-\sqrt{2}}{7}\right)\left(\frac{3+\sqrt{2}}{7}\right) = \frac{9 - 2}{49} = \frac{7}{49} = \frac{1}{7}\)
- Using the formula for a quadratic equation with roots \(x_1\) and \(x_2\):
\(x^2 - sx + p = 0\) - Substitute the sum and product of roots into the equation:
\(x^2 - \left(\frac{6}{7}\right)x + \frac{1}{7} = 0\) - Clear the fractions by multiplying the entire equation by 7 to obtain integer coefficients: \(7x^2 - 6x + 1 = 0\)
Thus, the quadratic equation whose roots are \(\frac{1}{3+\sqrt{2}}\) and \(\frac{1}{3-\sqrt{2}}\) is \(7x^2 - 6x + 1 = 0\).