If-P-x-x-3-ax-2-bx-c-with-a-b-and-c-real-numbers-Roots-of-P-x-z-3i-z-9i-and-2z-4-find-a-b-c-Note-z-is-complex-number-

Question Number 112478 by dw last updated on 08/Sep/20

I f P (x) = x^{3} + {a x}^{2} + b x + c, w i t h a, b a n d c r e a l n u m b e r s .

R o o t s o f P (x) z + 3 i, z + 9 i a n d 2 z - 4, f i n d ∣ a + b + c ∣ .

N o t e : z i s c o m p l e x n u m b e r .

Answered by floor(10²Eta[1]) last updated on 08/Sep/20

we know that a cubic polynomial with

real coefficients can have :

3 real roots or 1 real root and 2 complex roots

because if a polynomial has a complex

root, the conjugate of this complex number

is also a root .

If z \in R \Rightarrow 2 z - 4 \in R, but if z + 3 i and z + 9 i

are the complex roots so z - 3 i and z - 9 i

also should be . But {that}^{'} s impossible

because we just have 3 roots .

So z have an imaginary part \Rightarrow 2 z - 4

is one of the complex roots .

z = u + vi

roots : [2 u - 4 + 2 vi], [u + (3 + v) i], [u + (9 + v) i]

1 case :

u + (9 + v) i is the real root :

\Rightarrow 9 + v = 0 ∴ v = - 9

so (2 u - 4 - 18 i) is the conjugate of (u - 6 i)

but this is impossible because the

imaginary parts are different .

2 case :

u + (3 + v) i is the real root :

\Rightarrow 3 + v = 0 ∴ v = - 3

so (2 u - 4 - 6 i) is the conjugate of (u + 6 i)

\Rightarrow 2 u - 4 = u ∴ u = 4

roots : (4), (4 + 6 i), (4 - 6 i)

\Rightarrow P (x) = (x - 4) (x - 4 - 6 i) (x - 4 + 6 i)

P (1) = 1 + a + b + c

= - 3 (- 3 - 6 i) (- 3 + 6 i) = - 3.45

\Rightarrow a + b + c = - 135 - 1 = - 136

∣ a + b + c ∣= 136 .

Commented by MJS_new last updated on 08/Sep/20

good job!

Commented by dw last updated on 09/Sep/20

T h a n k y o u S i r

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