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Consider the equation (x 2 +x+1) 2 -(m-3)(x 2 +x+1) +m=0. Where m is a real parameter. The number of positive integral value of m for which two distinct real roots is??? Also The number of positive integral values of m ,m four distinct real roots is????

Consider the equation (x2+x+1)2-(m-3)(x2+x+1) +m=0. Where m is a real parameter. The number of positive integral value of m for which two distinct real roots is??? Also The number of positive integral values of m ,m
four distinct real roots is????

Grade:12

1 Answers

Satyajit Samal
19 Points
9 years ago
Suppose x^2+x+1 = t
So, the equation can be written as t^2 – (m-3)t + m = 0 
This equation will have two distinct real roots if 
(m-3)^2 > 4m
m^2 -10m + 9 > 0 
(m-1)(m-9) > 0 
m 9 …....(i)
 
Now t = x^2 + x + 1 = (x + 1/2)2+ 3/ 4
Part -1: So if t > 3/ 4 , then x2+ x + 1 = t will have two distinct roots, hence the original equation will have four distinct roots.
For the equation t2 - (m-3)t + m = 0 , if t > 3/ 4, then 
f (3/ 4) > 0 
9/ 16 – (m-3) 3/ 4 + m > 0 
m > -45/4 …......... (ii)
Also (m-3)/ 2 > 3/ 4 ( -b / 2a of quadratic is > 3/ 4)
m > 9/2 …............ (iii)
So, M should satisfy the conditions in (i), (ii) and (iii). 
 
Combining, all the positive integers > 9 ( 10, 11, 12, 13....) as values of m will give four distinct roots for the given equation. 
Part-2: For two real roots, both roots of equation  t2 - (m-3)t + m = 0 should be on different side of 3/ 4 , that is one greater than 3/ 4 , and the other one less than 3/ 4.
So, f (3/ 4)
m
Since we need positive integral values of m, no value for m possible in this case.

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