anonymous
  • anonymous
find a polynomial equation with integer coefficients, for which 5 is a double root and 1+ square root of 2 and 1- square root of 3i are solutions. I dont know how to solve this or write it out.. help?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
if A, B, C , D are roots of a polynomial, the the polynomial is the product of (x -A)(x-B)(x-C)(x-D) You have the four roots as 5, 5, 1 + root2 and 1-root 3i
anonymous
  • anonymous
k.then what do i do.
anonymous
  • anonymous
In the above equation, replace A, B, C and D by the given roots and take their product to get the polynomial

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anonymous
  • anonymous
so do (x-5)(x-5)(x-1+square root of 2)(x-1-square root of 3i)?
anonymous
  • anonymous
It will be (x-5)(x-5)[x-(1+root2)][x- (1- root of 3i)} Which becomes (x-5)(x-5)[x-1 - root2][x-1+ root of 3i}
anonymous
  • anonymous
ok
anonymous
  • anonymous
see if the co-efficients are integer and real then complex conjugate roots must exist. and if \[1+\sqrt{2}\] exists then \[1-\sqrt{2}\]also exits. let \[\sqrt{3i}=ki\] whre k is a no. \[(x-5)^{2}(x-1-\sqrt{2})(x-1+\sqrt{2})(x-1-ki)(x-1+ki)\]this gives \[(x-5)^{2}((x-1)^{2}-2)((x-1)^{2}+3)\] if you don't consider the conjugate part then you can't get integral coefficient....

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