just an easy practice pronlem

Algebra Level 3

If 3 + 3 i \large 3 + \sqrt{3} i is a root of the equation x 2 + p x + q = 0 \large x^2 + px + q = 0 , where p and q are real , then p + q = \large p + q =

This problem is a part of the set - 1's & 2's & QuEsTiOnS
none of these 6 5 ( 2 + i 3 ) 2 ( 2 + i \sqrt{3} )^2 10 18 16

Sakanksha Deo by Sakanksha Deo , 23, India

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3 solutions

Jason Hughes
Apr 27, 2015

x 2 + p x + q = 0 x^2+px+q=0 root 1 is 3 + 3 i 3+\sqrt3i the other root to have real coeffients is the complex conjugate 3 − 3 i 3-\sqrt3i .

x 2 + p x + q = ( x − ( 3 + 3 i ) ) ( x − ( 3 − 3 i ) ) x^2+px+q=(x-(3+\sqrt3i))(x-(3-\sqrt3i)) this yields that p = − ( 3 + 3 i ) − ( 3 − 3 i ) p= -(3+\sqrt3i)-(3-\sqrt3i) and q = ( 3 + 3 i ) ( 3 − 3 i ) q=(3+\sqrt3i)(3-\sqrt3i) p = − 6 , q = 12 , p + q = 6. p=-6,q=12, p+q=6.

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Edwin Gray
Sep 21, 2018

If 3 + sqrt(3(I is a root, then so is 3 - sqrt(3)I. Their sum = 6 and is equal to -p/1. Their product is 12 and equal to q/1. Therefore p + q = -6 + 12 = 6. Ed Gray

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Ramesh Goenka
Mar 24, 2015

Based on a simple concept that complex roots occur in pairs! The other root will be 3 - i sqrt(3) !

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