What is the monic quadratic equation with real coefficients having 2 + 3*i as one of its roots?
Answer and explanation
Correct answer: B. x^2 - 4*x + 13 = 0
- Step 1: Since coefficients are real, the roots must be complex conjugates: 2 + 3i and 2 - 3i.
- Step 2: Sum of roots = (2 + 3i) + (2 - 3i) = 4.
- Step 3: Product of roots = (2 + 3i)(2 - 3i) = 2^2 + 3^2 = 4 + 9 = 13.
- Step 4: The quadratic equation is x^2 - (sum)x + (product) = x^2 - 4x + 13 = 0.
Answer
x^2 - 4*x + 13 = 0.
Why not the others: x^2 + 4x + 13 = 0 has sum -4; x^2 - 4x - 5 = 0 computes 4 - 9 = -5 instead of 4 + 9; x^2 - 6*x + 13 = 0 uses twice the imaginary part for the sum.
Practise more Vectors, Analytic Geometry and Probability.
Free questions with an explanation for every one, and full timed Bagrut (Secondary School Matriculation) mock tests.
More Vectors, Analytic Geometry and Probability questions
- What is the modulus and principal argument of the complex number z = -1 + i*sqrt(3)?
- Compute the value of (1 + i)^10 using De Moivre's theorem.
- What is the sum of all complex solutions to the equation z^5 - 1 = 0?
- Find all four roots of the equation z^4 + 4 = 0 in the complex plane.
- Describe the geometric locus of points in the complex plane satisfying |z - (3 - 2*i)| = 5.
- Solve the complex equation z^2 + |z|^2 = 2 - 4i, where z = x + yi.