if alphe and beta are the zeros of the polynomial x square + 8x + 6, from the polynomial whose zeros are { question number (1) 1 by alpha square and 1 by beta square (2) 1 + beta by alphe and 1 + alphe by beta }
Answers
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Answer:
Step-by-step explanation:
x² + 8x + 6 = 0
α & β are roots
α + β = -8
αβ = 6
Equation whose roots area
1/α² & 1/β²
(x - 1/α²)(x-1/β²) = 0
= x² - x(1/α² + 1/β²) + 1/α²β² = 0
1/α² + 1/β² = (α² + β²)/(α²β²)
=>1/α² + 1/β² = ( (α + β)²- 2αβ)/((αβ)²)
Putting values of α + β = -8 & αβ = 6
=>1/α² + 1/β² = ( (-8)²- (2×6))/(6²)
=>1/α² + 1/β² = (64- 12)/(36)
=>1/α² + 1/β² = (52)/(36)
=>1/α² + 1/β² = (13)/(9)
x² - x(1/α² + 1/β²) + 1/α²β² = 0
=> x² - x(13/9) + 1/36 = 0
=> 36x² - 52x + 1 = 0
=> (6x)² - 52x + 1 = 0
Q2)
(x - (1+β)/α)(x- (1+α)/β)
= x² -x ((1+β)/α + (1+α)/β) )+ (1+β)(1+α)/αβ = 0
=> x² -x(β + β² + α + α²) + (1 + α² + β² + αβ) = 0
=> x² -x(α² + β² + α + β) + (1 + α² + β² + αβ) = 0
α + β = -8 , αβ = 6
α² + β² = (α + β)² - 2αβ
=> α² + β² = (-8)² - 2×6 = 52
Putting these values in
x² -x((α² + β² )+ (α + β)) + (1 + α² + β² + αβ) = 0
=> x² - x(52 -8) + (1 + 52 + 6) = 0
=> x² -44x + 59 = 0