The roots of 4x ^ 2 - 9x + a = 0 are sin beta and cos beta then a=?
Answers
Step-by-step explanation:
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Given α,β,γ are roots of equation
x
3
+px+q=0...(1)
then using properties of quadratic equation.
α+β+γ=0,αβγ=−q & αβ+βγ+γα=+p
then
y =
∣
∣
∣
∣
∣
∣
∣
∣
α
β
γ
β
γ
α
γ
α
β
∣
∣
∣
∣
∣
∣
∣
∣
=α(βγ−α
2
)−β(β
2
−γα)+γ(αβ−γ
2
)
⇒y=αβγ−α
3
−β
3
+αβγ−γ
3
+αβ
3
y=3αβγ=(α
3
+β
3
+γ
3
)...(2)
now (α+β+γ)
3
=α
3
+β
3
+γ
3
+3αβγ(αβ+βγ+γα).
or
(α+β+γ)
3
=(α
3
+β
3
+γ
3
)+3[(α+β+γ)(αβ+βγ+γα)−αβγ]
0=α
3
+β
3
+γ
3
+3[0+q]
[α
3
+β
3
+γ
3
=−3q]
From eq (2)
y=−3q−(−3q)=0
∵[y=0]
∴ value of given expression is zero.