if alpha and beta are zeros of polynomial f(x)=kx2+7x+7 such that alpha2+beta2=42 find k
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Answer:
Value of k is 1 or - 49/42.
Step-by-step explanation:
Given :
- α, β are zeroes of the polynomial f(x) = kx² + 7x + 7
- α² + β² = 42
Comparing the given polynomial with ax² + bx + c we get
- a = k
- b = 7
- c = 7
Sum of zeroes = α + β = - b/a = - 7/k
Product of zeroes = αβ = c/a = 7/k
We know that
( α + β )² = α² + β² + 2αβ
Substituting the known values
( - 7/k )² = 42 + 7/k
49/k² = 42 + 7/k
49/k² - 7/k = 42
( 49 - 7k ) / k² = 42
49 - 7k = 42k²
42k² + 7k - 49 = 0
42k² + 49k - 42k - 49 = 0
k( 42k + 49 ) - 1( 42k + 49 ) = 0
( k - 1 )( 42k + 49 ) = 0
k - 1 = 0 OR 42k + 49 = 0
k = 1 OR k = - 49/42
Therefore the value of k is 1 or - 49/42.
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