How can I solve this system of non-linear equations? { a + c = 0 b + a c + d = 6 b c + a d = − 5 b d = 6 The book I'm working from provides the answer a = − 1 , b = 1 , c = 1 , d = 6, but I'm having a hard time getting there.This is in order to reduce q ( x ) = x 4 + 6 x 2 − 5 x + 6 to a pair of irreducible quadratics ( x 2 + a x + b ) ( x 2 + c x + d ). We expand, gather like terms, and equate the coefficients of the resulting expression with those of the original, yielding the above system of equations.With some substitutions, I managed to get b 6 − 6 b 5 − 6 b 4 + 47 b 3 − 36 b 2 − 216 b + 216 = 0which I reduced to ( b − 6 ) ( b − 1 ) ( b 4 + b 3 − 5 b 2 + 6 b + 36 ) = 0Since the remaining factor ( b 4 + b 3 − 5 b 2 + 6 b + 36 ) has no real zeros, it also reduces to a pair of irreducible quadratics.

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How can I solve this system of non-linear equations?      {                            a          +          c                          =                          0                                      b          +          a          c          +          d                          =                          6                                      b          c          +          a          d                          =                          −          5                                      b          d                          =                          6                        The book I&#039;m working from provides the answer   a  =  −  1  ,  b  =  1  ,  c  =  1  ,  d  =  6, but I&#039;m having a hard time getting there.This is in order to reduce   q  (  x  )  =      x    4    +  6      x    2    −  5  x  +  6 to a pair of irreducible quadratics   (      x    2    +  a  x  +  b  )  (      x    2    +  c  x  +  d  ). We expand, gather like terms, and equate the coefficients of the resulting expression with those of the original, yielding the above system of equations.With some substitutions, I managed to get      b    6    −  6      b    5    −  6      b    4    +  47      b    3    −  36      b    2    −  216  b  +  216  =  0which I reduced to  (  b  −  6  )  (  b  −  1  )  (      b    4    +      b    3    −  5      b    2    +  6  b  +  36  )  =  0Since the remaining factor   (      b    4    +      b    3    −  5      b    2    +  6  b  +  36  ) has no real zeros, it also reduces to a pair of irreducible quadratics.

ejigaboo8y · Accepted Answer

You were so close! You noted that the quartic factor has no real zeros, so you can conclude that   b  =  1 or   b  =  6, whence   d  =  6 or   d  =  1 (respectively) by the last equation of your system. Since   c  =  −  a by the first equation of your system, then the third equation of your system becomes   a  (  d  −  b  )  =  −  5, which gives you   a, and thus   c  =  −  a.Both   (  a  ,  b  ,  c  ,  d  )  =  (  −  1  ,  1  ,  1  ,  6  ) and   (  a  ,  b  ,  c  ,  d  )  =  (  1  ,  6  ,  −  1  ,  1  ) work as solutions.

Lydia Carey · Accepted Answer

Do you know Gauss&#039;s lemma? A polynomial with integer coefficients will factorise over the rational numbers if and only if it will factorise over the integers. That is, your values for   a  ,  b  ,  c  ,  d must be integers.(  b  d  =  6, so for an easier solution you can just guess   b  =  ±  1  ,  ±  2  ,  ±  3  ,  ±  6 and see what works. Of course, by symmetry, if   b  =  −  1 then   d  =  −  6 and so on, so you only really need to try half of them.)

How can I solve this system of non-linear equations? { <mtable columnalign="left cen

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