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    how to Use b^2 - 4ac with a =3k, b= 8k + 6 and c = 9k- 2
    to get - 44k^2 +120k +36 > 0 as an answer....?
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    (Original post by riskanya)
    how to Use b^2 - 4ac with a =3k, b= 8k + 6 and c = 9k- 2
    to get - 44k^2 +120k +36 > 0 as an answer....?
    so you use b^2 - 4ac > 0 assuming two real roots

    (8k+6)^2 - 4 x 3k x (9k-2)
    (8k+6)(8k+6) -12k(9k-2) > 0
    64k^2 + 96k + 36 - 108k^2 + 24k > 0

    64 - 108 = -44k^2
    96 +24 = 120k
    therefore -44k^2 +120k +36 > 0
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    thank you
 
 
 
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