If beta and 1 /beta are zeroes of the polynomial (alpha 2 + alpha)x 2 + 61x + 6alpha, find values of beta and alpha
Yukeshkumar:
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Answered by
122
β and 1/β are the roots {zeros } of given polynomial (α² + α)x² +61x + 6α
Product of roots = constant/coefficient of x²
β.1/β = 6α/(α² + α)
1 = 6α/(α² + α)
α² + α - 6α = 0
α² - 5α = 0 ⇒ α = 0 and 5 , α ≠ 0
∴ α = 5
Sum of roots = - coefficient of x/coefficient of x²
β + 1/β = -61/ (α² + α) = -61/(5² + 5) = -61/30
30β² + 30β + 61 = 0
Here, Discriminant = (30)² - 4 × 30 × 61 < 0
So, there is no real roots of β
Hence, α = 5 and β ∉ R
Product of roots = constant/coefficient of x²
β.1/β = 6α/(α² + α)
1 = 6α/(α² + α)
α² + α - 6α = 0
α² - 5α = 0 ⇒ α = 0 and 5 , α ≠ 0
∴ α = 5
Sum of roots = - coefficient of x/coefficient of x²
β + 1/β = -61/ (α² + α) = -61/(5² + 5) = -61/30
30β² + 30β + 61 = 0
Here, Discriminant = (30)² - 4 × 30 × 61 < 0
So, there is no real roots of β
Hence, α = 5 and β ∉ R
Answered by
33
Answer:
Abhi, if you are reading this plz cconsider...
The answer is mostly right except that the last equation for Beta is 30beta^2+61beta+30..
That will give us 2 distinct roots
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