sum of the zeroes =(alpha+beta)=2+[-1/3]
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0
Answer:
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Answered by
1
Answer:
Given α+β=−8 & αβ=6
(α+β)
2
=α
2
+β
2
+2αβ
α
2
+β
2
=64−12=52
(α−β)
2
=α
2
+β
2
−2αβ
(α−β)
2
=52−12=40
(α−β)=2
10
now (α+β),(α−β) are roots of any quadratic equation,
quadratic equation is
[x−(α+β)][x−(α−β)]=0
(x+8)(x−2
10
)=0
x
2
+(8−2
10
)x−16
10
=0
Hence equation (1) is required equation
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