Question

The quadratic equation whose roots are \(\frac{1}{3+\sqrt{2}}\) and \(\frac{1}{3-\sqrt{2}}\), will be:

• A. \(7x^2 - 6x + 1 = 0\)
• B. \(6x^2 - 7x + 1 = 0\)
• C. \(x^2 - 6x + 7 = 0\)
• D. \(x^2 - 7x + 6 = 0\)

Hint:

Rationalize denominators before finding sum and product of roots.

Answer 1

The Correct Option is A

Concept: If roots are \(\alpha\) and \(\beta\), the quadratic is \(x^2 - (\alpha+\beta)x + \alpha\beta = 0\).

Step 1: Simplify the roots. \[ \frac{1}{3+\sqrt{2}} \cdot \frac{3-\sqrt{2}}{3-\sqrt{2}} = \frac{3-\sqrt{2}}{9-2} = \frac{3-\sqrt{2}}{7} \] \[ \frac{1}{3-\sqrt{2}} \cdot \frac{3+\sqrt{2}}{3+\sqrt{2}} = \frac{3+\sqrt{2}}{9-2} = \frac{3+\sqrt{2}}{7} \]

Step 2: Sum of roots. \[ \alpha + \beta = \frac{3-\sqrt{2}}{7} + \frac{3+\sqrt{2}}{7} = \frac{6}{7} \]

Step 3: Product of roots. \[ \alpha\beta = \frac{(3-\sqrt{2})(3+\sqrt{2})}{49} = \frac{9 - 2}{49} = \frac{7}{49} = \frac{1}{7} \]

Step 4: Form the quadratic. \[ x^2 - \frac{6}{7}x + \frac{1}{7} = 0 \quad \Rightarrow \quad 7x^2 - 6x + 1 = 0 \]