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Question

If α,β and γ are real numbers such that α2+β2+γ2=1 and α+β+γ=3, then β=
(correct answer + 1, wrong answer - 0.25)

A
13
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B
43
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C
23
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D
±13
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Solution

The correct option is A 13
α2+β2+γ2=1 (i)
α+β+γ=3 (ii)
From (ii), we have γ=3αβ (iii).
Substituting (iii) into (i):
α2+β2+(3αβ)2=1
α2+β2+3+α2+β223α23β+2αβ=1
α2+(β3)α+(β23β+1)=0

Since, α is real, D0
(β3)24(β23β+1)0
(3β1)20
β=13

Alternate Solution:
α2+β2+γ2=1 and α+β+γ=3
Let α=β=γ
3α2=1
α=±13 ...(1)
And 3α=3
α=13 ...(2)
From (1) and (2),
α=β=γ=13

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