if al and a2 are zeroes of polynomial kx^2 + 4x + 4 and if a1^2 + a2^2 = 24, find the value of k
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Answer:
2/3 or - 1
Step-by-step explanation:
a1 and a2 are the zeroes of kx² + 4x + 4
Comparing the given polynimial with ax² + bx + c we get,
- a = k
- b = 4
- c = 4
Sum of zeroes = a1 + a2 = - b / a = - 4 / k
Product of zeroes = a1.a2 = c / a = 4 / k
Given :
a1² + a2² = 24
Using algebraic identity ( a1 + a2 )² = a1² + a2² + 2a1.a2
⇒ ( - 4 / k )² = 24 + 2( 4 / k )
⇒ 16 / k² = 24 + 8 / k
Dividing by 8 on both sides
⇒ 2 / k² = 3 + 1 / k
⇒ 0 = 3 + 1 / k - 2 / k²
Multiplying by k² on sides
⇒ 0 = 3k² + k - 2
⇒ 3k² + k - 2 =0
⇒ 3k² + 3k - 2k - 2 = 0
⇒ 3k( k + 1 ) - 2( k + 1 ) = 0
⇒ ( 3k - 2 )( k + 1 ) = 0
⇒ 3k - 2 = 0 or k + 1 = 0
⇒ k = 2/3 or k = - 1
Therefore the value of k is 2/3 or - 1.
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