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If p+iq is the root of the eqn x^3+ax+b=0, then eqn whose one of the roots is 2p is?

If p+iq is the root of the eqn x^3+ax+b=0, then eqn whose one of the roots is 2p is?

Grade:11

2 Answers

Arun
25750 Points
4 years ago
It is given that, p+i q , is one of the roots of the equation 
       x³+ a x +b=0
This is a cubic equation.So, it has three roots.
As,Imaginary root occur in pairs.
 So, if , p + i q,is one root of the equation , then , p - i q, will be another root.It is given that third Root is equal to 2 p.
For, any Cubic equation of the form
   
If, u,v and w are roots of the equation
    
Then, ⇒⇒p+i q + p- i q + 2 p=0
         4 p=0
p=0
or
Also,⇒ (p +i q)(p-i q)×2 p= -b 
⇒(p²+q²)×2 p= -b
⇒2 p³+2 p q² + b=0
or
⇒(p+ i q)× (p -i q) + 2 p× ( p + i q)+2 p×(p- i q)=a
⇒p²+q²+2 p²+2 p i q+2 p² -2 p i q=a
⇒ 5 p² +q²-a=0
 
 
Honey
13 Points
one year ago
It's given That one of the root is p+iq let it be a
Then we know that the other root will be i.e b= p-iq
And the 3rd root be y
Hence, In the given equation x³+ax+b 
a+b+y =-b/a
p+iq+p-iq+ y= 0
2p+y=0 => -y=2p
Putting the required root 2p i.e -y we get
y3-ay +b = 0 the y variable can be written as in form of t,x,k,p or any other variables a/c to options. 
Hope it helps
 
 

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