Question 1099591
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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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<pre>
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^4 - 2x^3 + 4x^2 + 6x - 21}}} = 0,

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


{{{p^4 - 2*p^3 + 4*p^2 + 6*p - 21}}} = 0,                (1)     and

{{{(-p)^4 - 2*(-p)^3 + 4*(-p)^2 + 6*(-p) - 21}}} = 0,       (2),    or

{{{p^4 + 2*p^3 + 4*p^2 - 6p - 21}}} = 0.                 (3)


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

{{{4p^3 - 12p}}} = 0,    or  

{{{4p*(p^2-3)}}} = 0.


It implies that  p = +/- {{{sqrt(3)}}}.


Thus two roots of the given polynomial are  {{{sqrt(3)}}}  and  {{{-sqrt(3)}}}.


Then the polynomial is divisible by  {{{(x-sqrt(3))*(x+sqrt(3))}}} = {{{x^2-3}}}.


The quotient is  {{{x^2 -2x + 7}}}.


The roots of the last quadratic polynomial are  {{{1 + i*sqrt(6)}}}  and  {{{1 - i*sqrt(6)}}}.


Thus the roots of the original polynomial are  {{{sqrt(3)}}},  {{{-sqrt(3)}}},  {{{1 + i*sqrt(6)}}}  and  {{{1 - i*sqrt(6)}}}.
</pre>

Solved.