The product of two of the four roots of the equation x⁴–18x³+Kx²+200x–1984=0 is –32. Find the value of K.

The solution of the puzzle is given below.
Since the product of two of the roots of the given equation is -32; it follows that, the product of the remaining roots is equal to  -1984/ -32 = 62. Accordingly, we can write the given quartic expression as the product of two quadratic expressions, so that:
 x^4 – 18x^3 + K*x^2 + 200x – 1984

= (x^2 + px -32)(x^2 + qx + 62);(say)
Comparing the exponents of x, x^2 and x^3 in the above relationship, we obtain:
(I) p+q = -18; (II) K = 30 + pq; and (III) 62*p – 32*q = 200;
Solving  for p and q in  (I) and (III), we obtain, (p,q) = (-4,-14); giving :
K= 30 + (-4)*(-14) = 86.
Note: For a harder variation of quartic methodologies, refer to  "A Quartic Problem"
( Reference: http://perplexus.info/show.php?pid=4279).

Edited on May 15, 2006, 12:56 pm

Edited on May 15, 2006, 1:00 pm

Posted by K Sengupta on 2006-05-15 12:54:08