Simplex!
Given a monic quadratic a x 2 + b x + c with one of its roots equal to 4 − 3 i , Find b + c
Note- a , b , c indicate the coefficients of the quadratic. A monic polynomial has the leading coefficient as 1 . Thus, here- a = 1 . i denotes the square root of − 1
The answer is 17.
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2 solutions
Except you meant ( 3 i ) 2 , not 3 i 2 .
I have simply put 4-3i into the equation and solved for b and c.
Just for fun Does monic has anything to do with harmonic
The equation is x 2 + b x + c .
One of the root is 4 − 3 i ,implies other root is 4 + 3 i .
Sum of the roots = 1 − b = 8 → b = − 8
Product of the roots is 1 c = ( 4 + 3 i ) ( 4 − 3 i ) = 4 2 − ( 3 i ) 2 = 1 6 + 9 = 2 5 .
Therefore b + c = − 8 + 2 5 = 1 7
Given- Its a monic and also has one root= 4 − 3 i . Thus by complex conjugate root theorem- 4 + 3 i must also be a root. Thus all we have to do now is- Vieta! Yes, sum of the two roots= 8 and product= 4 2 − ( 3 i ) 2 = 2 5 . Thus b + c = − 8 + 2 5 = 1 7