To form a polynomial equation with rational coefficients whose roots include $\sqrt{2} + \sqrt{3}$, we must use the fact that the roots must come in conjugate pairs if the coefficients are to remain rational. Therefore, the conjugate root $\sqrt{2} - \sqrt{3}$ must also be included.

Let's find a polynomial with these roots:

Step 1: Define the roots

Let the roots be: $x_1 = \sqrt{2} + \sqrt{3} \quad \text{and} \quad x_2 = \sqrt{2} - \sqrt{3}$

Step 2: Form the quadratic equation

The polynomial can be expressed as: $(x - x_1)(x - x_2) = 0$

Expanding this: $(x - (\sqrt{2} + \sqrt{3}))(x - (\sqrt{2} - \sqrt{3})) = 0$

Step 3: Expand the expression

We expand the product: $x^2 - x(\sqrt{2} + \sqrt{3}) - x(\sqrt{2} - \sqrt{3}) + (\sqrt{2} + \sqrt{3})(\sqrt{2} - \sqrt{3})$

This simplifies as follows:

• The middle terms $-x(\sqrt{2} + \sqrt{3}) - x(\sqrt{2} - \sqrt{3})$ cancel out since $(\sqrt{2} + \sqrt{3}) + (\sqrt{2} - \sqrt{3}) = 2\sqrt{2}$, leaving: $x^2 - \left((\sqrt{2})^2 - (\sqrt{3})^2\right)$

• The last term simplifies using the difference of squares: $(\sqrt{2} + \sqrt{3})(\sqrt{2} - \sqrt{3}) = (\sqrt{2})^2 - (\sqrt{3})^2 = 2 - 3 = -1$

So, the polynomial is: $x^2 - (\text{expression from the last step}) = x^2 - (-1) = x^2 + 1$

Thus, the polynomial equation with rational coefficients that has $\sqrt{2} + \sqrt{3}$ as a root is: $\boxed{x^2 - 2\sqrt{2}x - 1 = 0}$

Would you like further clarification on any of these steps, or do you have any other questions?

Relative Questions

1. How can you derive a polynomial equation for any irrational root?
2. What is the significance of conjugate pairs in polynomial equations with rational coefficients?
3. How do you expand and simplify expressions involving square roots?
4. Can you find other types of polynomials with roots involving both square and cube roots?
5. What is the general method to factor a polynomial if one of its roots is known?

Tip

Always check for conjugate pairs when forming polynomials with irrational or complex roots, as they ensure the coefficients remain rational.