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"
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"
✨ Free AI Tools for You
Powered by Chatterbot AI
Answer
To find the quadratic equation with rational coefficients, given that one of the roots is (3 + \sqrt{2}), follow these steps:

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

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: $$ x^2  (\alpha + \beta)x + \alpha \beta = 0 $$

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: $$ \alpha \beta = (3 + \sqrt{2})(3  \sqrt{2}) = 3^2  (\sqrt{2})^2 = 9  2 = 7 $$

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.
India's 1^{st} AI Doubt Solver for CBSE, JEE, and NEET
Ask a Question for Freeand then it's just ₹212 a month