The question asks about a cubic equation $x^3 + ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are real numbers. One of the roots of this cubic equation is given as $-1 + \sqrt{2}i$.

Steps to solve:

1. Since the coefficients of the cubic equation are real, the complex conjugate of the given root $-1 + \sqrt{2}i$, which is $-1 - \sqrt{2}i$, must also be a root of the equation.

2. Thus, the three roots of the cubic equation are: $-1 + \sqrt{2}i, \quad -1 - \sqrt{2}i, \quad \text{and a real root} \quad r.$ Let's represent the real root as $r$.

(1) Express $b$ and $c$ in terms of $a$:

• From Vieta’s formulas, we know the following relations hold for a cubic equation:

  • The sum of the roots: $(-1 + \sqrt{2}i) + (-1 - \sqrt{2}i) + r = -a$ Simplifying this: $-2 + r = -a \quad \Rightarrow \quad r = a - 2.$

  • The sum of the product of the roots taken two at a time: $(-1 + \sqrt{2}i)(-1 - \sqrt{2}i) + (-1 + \sqrt{2}i)r + (-1 - \sqrt{2}i)r = b.$

    The product of $(-1 + \sqrt{2}i)$ and $(-1 - \sqrt{2}i)$ is: $(-1 + \sqrt{2}i)(-1 - \sqrt{2}i) = (-1)^2 - (\sqrt{2}i)^2 = 1 - (-2) = 3.$ Therefore, the equation for $b$ becomes: $3 + (-1 + \sqrt{2}i)r + (-1 - \sqrt{2}i)r = b.$ Simplifying: $3 + (-1 + \sqrt{2}i + -1 - \sqrt{2}i)r = b,$ $3 + (-2)r = b \quad \Rightarrow \quad b = 3 - 2r.$ Substituting $r = a - 2$ into this: $b = 3 - 2(a - 2) = 3 - 2a + 4 = 7 - 2a.$

(2) Express the real root of the equation in terms of $a$:

From the sum of the roots, we already found that $r = a - 2$.

Final answers:

1. $b = 7 - 2a$
2. The real root of the equation is $a - 2$.

Would you like more details, or do you have any questions on this solution?

5 Related Questions:

1. What is Vieta's formula and how is it used in polynomial equations?
2. How do you find the product of two complex conjugates?
3. Can all cubic equations with real coefficients have complex roots?
4. How do real and complex roots affect the shape of the graph of a cubic equation?
5. How would the equation change if the real root were a specific value, say $r = 0$?

Tip:

For any polynomial with real coefficients, complex roots always appear in conjugate pairs, ensuring the result remains real.