InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The equation of smallest degree with real co-efficients having 2 + 3i as one of the roots is : (a) x2 − 4x + 13 = 0 (b) x2 + 5x + 6 = 0 (c) x2 − 2x + 1 = 0 (d) x2 + 2x + 1 = 0 |
| Answer» <p style="text-align:justify"><span style="font-family:Arial,Helvetica,sans-serif"><span style="color:#000000"><strong>Correct option (a) x<sup>2</sup> − 4x + 13 = 0</strong></span></span></p><p style="text-align:justify"><span style="font-family:Arial,Helvetica,sans-serif"><span style="color:#000000"><strong>Explanation:</strong></span></span></p><p style="text-align:justify"><span style="font-family:Arial,Helvetica,sans-serif"><span style="color:#000000">Since 2 + 3i is a root </span></span></p><p style="text-align:justify"><span style="font-family:Arial,Helvetica,sans-serif"><span style="color:#000000">∴ 2 − 3i is also a root Hence, required equation is </span></span></p><p style="text-align:justify"><span style="font-family:Arial,Helvetica,sans-serif"><span style="color:#000000">x<sup>2</sup> − (sum of roots) x + (product of roots) = 0 </span></span></p><p style="text-align:justify"><span style="font-family:Arial,Helvetica,sans-serif"><span style="color:#000000">Sum of roots = 2 + 3i + 2 − 3i = 4 </span></span></p><p style="text-align:justify"><span style="font-family:Arial,Helvetica,sans-serif"><span style="color:#000000">Product of roots </span></span></p><p style="text-align:justify"><span style="font-family:Arial,Helvetica,sans-serif"><span style="color:#000000">= (2 + 3i) (2 − 3i) = 4 − 9i<sup>2 </sup>= 13 </span></span></p><p style="text-align:justify"><span style="font-family:Arial,Helvetica,sans-serif"><span style="color:#000000">∴ [ i<sup>2</sup> = −1]</span></span></p><p style="text-align:justify"><span style="font-family:Arial,Helvetica,sans-serif"><span style="color:#000000">So, equation is x<sup>2</sup> − 4x + 13 = 0.</span></span></p> | |