Q.
Let (x1,x2),(x2,x3) and (x3,x1) are respectively the roots of x2−2ax+2=0,x2−2bx+3=0 and x2−2cx+6=0, where x1,x2,x3>0. Find the value of (a+b+c).
60
86
Complex Numbers and Quadratic Equations
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Answer: 6
Solution:
Clearly, x1x2=2,x2x3=3 and x3x1=6 ⇒(x1x2)(x2x3)(x3x1)=36 or x1x2x3=6 (As, x1,x2,x3 are all positive.)
So, x1=x2x3x1x2x3=36=2
Similarly, x2=x1x3x1x2x3=66=1
and x3=x1x2x1x2x3=26=3
Also, we have 2a=x1+x2=3;2b=x2+x3=4 and 2c=x3+x1=5
Hence, 2(a+b+c)=12. ⇒(a+b+c)=6.