maths

Asked by kanishkaaggarwal | 21st Feb, 2010, 12:45: PM

Expert Answer:

ax2 +bx +c = 0

x2 +bx/a + c/a = 0

α+β = -b/a and αβ = c/a

Now

α' = α+3 and β' = β+3

(α+3) + (β +3) = -b/a + 6 = (6a-b)/a = α' + β'

and

(α+3-3)(β+3-3) = c/a

(α'-3)(β'-3) = c/a

-3(α' + β') + α' β' + 9 = c/a

α' β' = c/a - 9 +3(α' + β')

= c/a - 9 +3((6a-b)/a)

= c/a - 9 + (18a-3b)/a

= (c - 9a + 18a - 3b)/a

= (9a - 3b + c)/a

Hence the equation with roots α' = α+3 and β' = β+3,

x2 + (6a-b)x/a + (9a - 3b + c)/a = 0

ax2 + (6a-b)x + (9a - 3b + c) = 0

Regards,

Team,

TopperLearning.

 

 

Answered by  | 21st Feb, 2010, 07:12: PM

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