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x² - 6x + k = 0
here
α+β = -(-6)/1 = 6
αβ = k/1 = k
α² + β² = (α +β)² - 2αβ = 40
⇒ 6² - 2k = 40
⇒ - 2k = 40 - 36 = 4
k = 4 / -2
⇒ k = -2
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here
α+β = -(-6)/1 = 6
αβ = k/1 = k
α² + β² = (α +β)² - 2αβ = 40
⇒ 6² - 2k = 40
⇒ - 2k = 40 - 36 = 4
k = 4 / -2
⇒ k = -2
.... I hope it will help you !!!
.... Please mark it as brainliest !!!
Answered by
1
Given α² + β² = 40
(By formula (a+b)² = a² + b² + 2ab
a² + b² = (a+b)² - 2ab)
Then, α² + β² = 40
(α + β)² - 2αβ = 40 ..........(1)
In, p(x) = x² - 6x + k
Sum of zeroes = α + β
α + β = -b/a
= -(-6)/1 = 6 ..........(2)
Product of zeroes = αβ
αβ = c/a
= k/1 = k ...........(3)
From (1) (2) (3) substituting values,
(α + β)² - 2αβ = 40
(6)² - 2(k) = 40
-2k = 40 - 36
k = 4/-2 = -2
Hence k = -2
(By formula (a+b)² = a² + b² + 2ab
a² + b² = (a+b)² - 2ab)
Then, α² + β² = 40
(α + β)² - 2αβ = 40 ..........(1)
In, p(x) = x² - 6x + k
Sum of zeroes = α + β
α + β = -b/a
= -(-6)/1 = 6 ..........(2)
Product of zeroes = αβ
αβ = c/a
= k/1 = k ...........(3)
From (1) (2) (3) substituting values,
(α + β)² - 2αβ = 40
(6)² - 2(k) = 40
-2k = 40 - 36
k = 4/-2 = -2
Hence k = -2
☺ Hope this Helps ☺
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