CBSE Class 10 Answered
maths
Asked by kanishkaaggarwal | 21 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 | 21 Feb, 2010, 07:12: PM
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