Question

Let a=2-3i be a root of the equation z²-4z+k=0 , where k is a real number. If β is the other root, then the value of α² +β² is

• A. 26
• B. -5
• C. 5
• D. 10
• E. -10

Answer 1

Answer: (E)

We are given that \(\alpha = 2 - 3i\) is a root of the quadratic equation: 

\(z^2 - 4z + k = 0\), where \(k\) is a real number. Let \(\beta\) be the other root, and we need to find \(\alpha^2 + \beta^2\).

From Vieta's formulas, for the equation \(z^2 - 4z + k = 0\), we know the following:

• \(\alpha + \beta = 4\) (sum of the roots)
• \(\alpha \beta = k\) (product of the roots)

Since \(\alpha = 2 - 3i\), we can calculate \(\alpha^2\):

\(\alpha^2 = (2 - 3i)^2 = 4 - 12i + 9i^2 = 4 - 12i - 9 = -5 - 12i\)

Next, using \(\alpha + \beta = 4\), we get:

\(\beta = 4 - \alpha = 4 - (2 - 3i) = 2 + 3i\)

Now, calculate \(\beta^2\):

\(\beta^2 = (2 + 3i)^2 = 4 + 12i + 9i^2 = 4 + 12i - 9 = -5 + 12i\)

Finally, calculate \(\alpha^2 + \beta^2\):

\(\alpha^2 + \beta^2 = (-5 - 12i) + (-5 + 12i) = -5 - 5 = -10\)

The answer is -10.