If alpha and beta are the zeros of th polynomial f(x) =x^2-k(x+1)-a, and if (alpha+1)(beta+1)=0;find A
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Heya User,
*Quadratic Reduction* ---> Using Zero-Co-efficient Relation:->
---> f(x) = x² - k( x + 1 ) - a = x² - kx - ( k + a )
---> Let α & β be the 2 zeroes -->
Then,
--> [ α + β ] = -(-k)
--> αβ = -(k+a)
Hence, --> putting the values in (α + 1)(β + 1) = 0, we get:->
--> [ αβ + α + β + 1 ] = 0
==> [ -(k+a) + k + 1 ] = 0
==> [ -k - a + k + 1 ] = 0
==> [ k - k + 1 - a ] = 0
==> a = 1 ...
Further, --> for discriminant -->
---> b² - 4ac ≤ 0
==> k² - 4(-k-a) ≤ 0
==> k² + 4k + 4 ≤ 0
==> [ k + 2 ]² ≤ 0
==> But then, the only real soln would be --> k = -2..
Hence, the values of k-->(-2) and a=1 holds for the given eqn.
*Quadratic Reduction* ---> Using Zero-Co-efficient Relation:->
---> f(x) = x² - k( x + 1 ) - a = x² - kx - ( k + a )
---> Let α & β be the 2 zeroes -->
Then,
--> [ α + β ] = -(-k)
--> αβ = -(k+a)
Hence, --> putting the values in (α + 1)(β + 1) = 0, we get:->
--> [ αβ + α + β + 1 ] = 0
==> [ -(k+a) + k + 1 ] = 0
==> [ -k - a + k + 1 ] = 0
==> [ k - k + 1 - a ] = 0
==> a = 1 ...
Further, --> for discriminant -->
---> b² - 4ac ≤ 0
==> k² - 4(-k-a) ≤ 0
==> k² + 4k + 4 ≤ 0
==> [ k + 2 ]² ≤ 0
==> But then, the only real soln would be --> k = -2..
Hence, the values of k-->(-2) and a=1 holds for the given eqn.
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