SOLUTION: prove that the equation mx(x^2+2x+3)=x^2-2x-3 has exactly one real root if m=1 and exactly 3 real roots if m=-2/3

Algebra ->  Trigonometry-basics -> SOLUTION: prove that the equation mx(x^2+2x+3)=x^2-2x-3 has exactly one real root if m=1 and exactly 3 real roots if m=-2/3      Log On


   

Question 977117: prove that the equation mx(x^2+2x+3)=x^2-2x-3 has exactly one real root if m=1 and exactly 3 real roots if m=-2/3
Answer by anand429(138) About Me  (Show Source):
You can put this solution on YOUR website!
Simplifying, our equation becomes,
mx%5E3+%2B+%282m-1%29x%5E2+%2B+%283m%2B2%29x+%2B+3+=0
For m =1,
we get,
x%5E3+%2B+x%5E2+%2B+5x+%2B+3+=0
Comparing with standard form,
ax%5E3%2Bbx%5E2%2Bcx%2Bd+=+0
Let f+=+%28%283c%2Fa%29-%28b%5E2%2Fa%5E2%29%29%2F3
=> f=+%2815+-1%29%2F3
=> f+=+14%2F3
Let g=+%28%282b%5E3%2Fa%5E3%29-%289bc%2Fa%5E2%29+%2B+%2827d%2Fa%29%29%2F27
=> g=+%282-45%2B81%29%2F27
=> g=38%2F27
Now, let h=+%28g%5E2%2F4%29+%2B+%28f%5E3%2F27%29
=> h=+%2838%2F27%29%5E2%2F4+%2B+%2814%2F3%29%5E3%2F27
Clearly, h>0
hence, there is only one real root.
Similarly, put m = -2/3 and solve as above,
If h<0, it means all three roots are real.