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Q.

Let a, b, c be the lengths of three sides of a triangle satisfying the condition (a2+b2)x22b(a+c)x+(b2+c2)=0. If the set of all possible values of ‘x’ is the interval (α,  β), then 24(α2+β2) is equal to ……… 

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a

72

b

96

c

36

d

84

answer is B.

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Detailed Solution

(a2+b2)x22b(a+c)x+(b2+c2)=0 
(ax+b)2+(bxc)2=0  {x=ba=cbb=ax,  c=bx=ax2 
now,  a+b>c  and  ab<c 
1+x>x2  and  1x<x2 
x(512,  5+12)=(α,  β)  
24(α2+β2)=72

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Let a, b, c be the lengths of three sides of a triangle satisfying the condition (a2+b2)x2−2b(a+c)x+(b2+c2)=0. If the set of all possible values of ‘x’ is the interval (α,  β), then 24(α2+β2) is equal to ………