Question

The quadratic equation with rational coefficients whose one root is $3+\sqrt{2}$ is:

A $x^2 + 2x + 1 = 0$ B $x^2 - 6x - 7 = 0$ C $x^2 - 6x + 7 = 0$ D $x^2 + 6x - 7 = 0"

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Answer

To find the quadratic equation with rational coefficients, given that one of the roots is (3 + \sqrt{2}), follow these steps:

  1. Identify the given root: One root is (3 + \sqrt{2}).

  2. Understand the nature of roots with rational coefficients:

    • If a quadratic equation has rational coefficients and one of its roots is irrational (involving a square root), the other root must be its conjugate. This is to ensure the equation has rational coefficients when expanded.
    • Therefore, if one root is (3 + \sqrt{2}), the other root must be (3 - \sqrt{2}).
  3. Form the quadratic equation:

    • Let the roots be (\alpha = 3 + \sqrt{2}) and (\beta = 3 - \sqrt{2}).
    • The standard form of a quadratic equation with roots (\alpha) and (\beta) is: $$ x^2 - (\alpha + \beta)x + \alpha \beta = 0 $$
  4. Calculate the sum and product of the roots:

    • Sum of the roots: $$ \alpha + \beta = (3 + \sqrt{2}) + (3 - \sqrt{2}) = 6 $$
    • Product of the roots: $$ \alpha \beta = (3 + \sqrt{2})(3 - \sqrt{2}) = 3^2 - (\sqrt{2})^2 = 9 - 2 = 7 $$
  5. Substitute the values into the standard equation form:

    • Using the sum (\alpha + \beta = 6) and the product (\alpha \beta = 7), we get: $$ x^2 - 6x + 7 = 0 $$

Therefore, the quadratic equation with rational coefficients whose one root is (3 + \sqrt{2}) is: (\boxed{x^2 - 6x + 7 = 0}).

Thus, the correct answer is Option C.


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