Question

Alpha and beta are the zeroes of kx2 4x 4 alpha2 beta2 ]= 24 then find k

Answer 1

Some error in your question , question is in such a way --> α and β are the zero of the Kx² + 4x + 4 , α² + β² = 24 then find k ?

Solution :- α and β are the zeros of the given polynomial Kx² + 4x + 4 = 0

so, product of zeros = αβ = constant/coefficient of x² = 4/K

sum of zeros = α + β = -coefficient of x/Coefficient of x² = -4/k

Now, α² + β² = 24

⇒(α + β)² - 2αβ = 24

⇒(-4/k)² - 2(4/k) = 24

⇒16/K² - 8/k = 24

⇒ 2 - k = 3k²

⇒3k² + k -2 = 0

⇒ 3k² + 3k - 2k - 2 = 0

⇒3k(k + 1) - 2(k +1) = 0

⇒(3k -2)(k + 1) = 0

Hence, k = 2/3 and -1

Answer 2

Answer:

According to the question,α and β are the zeroes of polynomial kx² + 4x + 4a

= k-b

= -4c = 4α + β = -b/a

= -4/kαβ = c/a = 4/k

Given = α² + β² = 24

(α + β)² - 2αβ = 24=

(-4/k)² - 2×4/k = 24

= 16/k² - 8/k = 24

= 16 - 8k = 24k²

= 24k² + 8k - 16 = 0

= 3k² + k - 2 = 0

= 3k² + 3k - 2k - 2 = 0

= 3k(k + 1) - 2(k + 1) = 0

= (3k - 2)(k + 1) = 0

= k = 2/3 or -1