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If α, β, γ are zeroes of cubic polynomial x³ + px² + qx + 2 such that αβ+1=0. Find the value of 2p+q+5.
Solutions.
Here We have,
p(x) = x³ + px² + qx + 2
αβ+1=0
(Given)
αβ =
-1 …..eq(i)
We know that if α, β, γ are zeroes of any
cubic polynomial, then
αβ+βγ+γα = c/a = q/1
-1+
βγ+γα = q (from eq(i) αβ = -1)
βγ+γα =
q +1
γ(α+β)=
q +1 …..eq(ii)
Product of zeroes of cubic polynomial is -d/a
αβγ =
-2 (from eq(i) αβ = -1)
(-1)γ =
-2
γ =
2 …..eq(iii)
Now, Putting the value of γ from eq(iii) in
eq(ii) we get
2(α+β)=
q +1
α+β =
(q +1)/2 …..eq(iv)
We know that the sum of zeroes is -b/a
α+β+γ =
-p/1 (α+β = (q +1)/2 and γ = 2 )
(q +1)/2 +2 = -p
(q +1+4)/2 = -p
q + 5 =
-2p
2p + q
+ 5 = 0
Thus , the value of 2p+q+5 is 0.
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