SOLUTION: if alpha,beta,gamma,and delta(symbols) are the roots of x^4-2x^3+4x^2+6x-21=0 if alpha + beta =0, solve the equation completely.

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: if alpha,beta,gamma,and delta(symbols) are the roots of x^4-2x^3+4x^2+6x-21=0 if alpha + beta =0, solve the equation completely.      Log On


   



Question 1099591: if alpha,beta,gamma,and delta(symbols) are the roots of x^4-2x^3+4x^2+6x-21=0 if alpha + beta =0, solve the equation completely.
Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
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if alpha,beta,gamma,and delta(symbols) are the roots of x^4-2x^3+4x^2+6x-21=0 if alpha + beta =0, solve the equation completely.
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I will use  "p"  and  "q"  instead of alpha and beta to make my writing easier.


So, I am given that p and q are the roots of the equation 

x%5E4+-+2x%5E3+%2B+4x%5E2+%2B+6x+-+21 = 0,

such that  p = -q.  Then the fact that p and q are the roots means


p%5E4+-+2%2Ap%5E3+%2B+4%2Ap%5E2+%2B+6%2Ap+-+21 = 0,                (1)     and

%28-p%29%5E4+-+2%2A%28-p%29%5E3+%2B+4%2A%28-p%29%5E2+%2B+6%2A%28-p%29+-+21 = 0,       (2),    or

p%5E4+%2B+2%2Ap%5E3+%2B+4%2Ap%5E2+-+6p+-+21 = 0.                 (3)


Now subtract eq(1) from eq(3) (both sides). You will get

4p%5E3+-+12p = 0,    or  

4p%2A%28p%5E2-3%29 = 0.


It implies that  p = +/- sqrt%283%29.


Thus two roots of the given polynomial are  sqrt%283%29  and  -sqrt%283%29.


Then the polynomial is divisible by  %28x-sqrt%283%29%29%2A%28x%2Bsqrt%283%29%29 = x%5E2-3.


The quotient is  x%5E2+-2x+%2B+7.


The roots of the last quadratic polynomial are  1+%2B+i%2Asqrt%286%29  and  1+-+i%2Asqrt%286%29.


Thus the roots of the original polynomial are  sqrt%283%29,  -sqrt%283%29,  1+%2B+i%2Asqrt%286%29  and  1+-+i%2Asqrt%286%29.

Solved.