Topic:Theory of Equations
Problem of 2011-February-15
Problem: Let m be a real number such that a and b are roots of the equation x
2+(m-4)x+m
2-3m+3=0
are real.
Find the value of m for which a
2+b
2=6
Options:
A. -√5-1
B. √5+1
C. √5-1
D. None of these
Solution:
Since a and b are real so D ≥0
=> (m-4)2 + -4(m2-3m+3) = -3m2+4m+4 ≥ 0
=> -2/3 ≤ m ≤2
Also a2+b2 = 6
=> (a+b)2 - 2ab = 6 => -m2-2m+10=0
We get m =-1±5
Now -1-√5<-2/3<-1+√5<2
So m = √5-1
Hence option C is correct.