SOLUTION: if the roots of the equation x^2+x+m=0 be real and unequal.Prove that the roots of the equation 2x^2+4(1+m)x+2m^2+3=0 are imaginary. where m is real.

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: if the roots of the equation x^2+x+m=0 be real and unequal.Prove that the roots of the equation 2x^2+4(1+m)x+2m^2+3=0 are imaginary. where m is real.      Log On


   



Question 1095554: if the roots of the equation x^2+x+m=0 be real and unequal.Prove that the roots of the equation 2x^2+4(1+m)x+2m^2+3=0 are imaginary. where m is real.
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
...the roots of the equation x^2+x+m=0 be real and unequal.
So the dicriminant b%5E2-4ac=%281%29%5E2-4%281%29m=1-4m%22%22%3E%22%220
                                         -4m%22%22%3E%22%22-1
                                           m%22%22%3C%22%221%2F4

Prove that the roots of the equation 2x^2+4(1+m)x+2m^2+3=0 are imaginary. where m is real.
 
2x%5E2%2B4%281%2Bm%29x%2B2m%5E2%2B3%22%22=%22%220

We have to prove that the discriminant of that is negative:

discriminant%22%22=%22%22b%5E2-4ac%22%22=%22%22%284%281%2Bm%29%5E%22%22%29%5E2-4%282%29%282m%5E2%2B3%29%22%22=%22%22

16%281%2Bm%29%5E2+-+8%282+m%5E2+%2B+3%29%22%22=%22%2216%281%2B2m%2Bm%5E2%29-16m%5E2-24%22%22=%22%22

16%2B32m%2B16m%5E2-16m%5E2-24%22%22=%22%2232m-8

Since m%22%22%3C%22%221%2F4
      32m%22%22%3C%22%228
      32m-8%22%22%3C%22%228-8
      32m-8%22%22%3C%22%220

Proved.

Edwin