What are the roots of the quadratic equation 2 gamma^ 2 +6=-7 gamma. ?
Answers
Answer:
As α,β,γ are roots of x
3
+27=0
s
1
=α+β+γ=0⇒β+γ=−α
s
2
=αβ+βγ+γα=0⇒α(γ+β)+βγ=0⇒α
2
=γβ
s
3
=αβγ=27
Now,
(α+β+γ)
2
=α
2
+β
2
+γ
2
+2(αβ+βγ+γα)
⇒α
2
+β
2
+γ
2
=0
Therefore,
S
1
=(
α
γ
)
2
+(
α
β
)
2
=
α
2
γ
2
+β
2
=
α
2
−α
2
=−1
S
2
=(
α
γ
)
2
(
α
β
)
2
=(
α
2
γβ
)
2
=(
α
2
α
2
)
2
=1
Hence required equation is
x
2
−S
1
x+S
2
=0
⇒x
2
+x+1
NOTE :
I am putting x instead of gamma here. so solve accordingly in your notebook
Step-by-step explanation: now here is your answer
2x^2 + 6 = 7x
2x^2 - 7x + 6 = 0
2x^2 - 4x - 3x + 6 = 0
2x (x - 2) - 3 (x - 2) = 0
(2x - 3) (x - 2) = 0
2x - 3 = 0
2x = 3
x = 3/2
AND {you have to right both because a quadratic equation
always has two roots exception where both the roots
are equal}
x - 2 = 0
x = 2