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if 2x^2+6x+b is a quadratic equation, where b is less than zero. the the value of alpha/beta + beta/alphs is less than

if 2x^2+6x+b is a quadratic equation, where b is less than zero. the the value of alpha/beta + beta/alphs is less than
 

Grade:12

5 Answers

Harsh Patodia IIT Roorkee
askIITians Faculty 907 Points
9 years ago
i assume alpha and beta are roots of above equation

sum of roots = -3
Product of roots= b/2

\alpha /\beta +\beta /\alpha = (\alpha ^{2} +\beta ^{2})/\alpha \beta
=((\alpha +\beta )^{^{2}}- 2\alpha \beta )/\alpha \beta
= (9-b/2)/b/2
= (18-b)/b = 18/b -1
Since b is less than 0 first term will be negative
So maximum value of above will be -1.
vinod
22 Points
9 years ago
I understand answer is less than 2
vinod
22 Points
9 years ago
Sorry, answer is less than -2. 
Harsh Patodia IIT Roorkee
askIITians Faculty 907 Points
9 years ago
Yes i miss that part
Answer will be -2
At this step
= [(9-2(b/2))/b/2]
= (18-2b)/b = 18/b -2
Since b is less than 0 first term will be negative
So maximum value of above will be -2.
OM Prakash Beniwal
21 Points
6 years ago
Yes, answer will be – 2. On putting values we get 
α/β + β/α  = (α2 + β2 )/α β = [ (α + β )2  – 2α β ] /α β 
[ (– 3)2 – 2 b/2]/ (b/2) = (9–  b) / (b/2) = 18/b – 2
Since b
If b is numerically very large, 18/b will be very small negative (i.e., very close to 0, but less than 0).
Thus maximum value of α/β + β/α   – 2.
 
 

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