please answer tomorrow is my test no 17
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( k + 1 ) x² + 2 ( k + 3 ) x + ( k + 8 ) = 0
Discriminant
Δ = [ 2(k+3) ]² - 4(k+1)(k+8)
= 4(k+3)² - 4(k+1)(k+8)
= 4 [ ( k² + 6k + 9 ) - ( k² + 9k + 8 ) ]
= 4 ( 1 - 3k )
Solutions:
x = [ - 2 ( k + 3 ) ± √( 4 ( 1 - 3k ) ) ] / [ 2 ( k + 1 ) ]
= [ -k - 3 ± √( 1 - 3k ) ] / ( k + 1 )
Note, there are only real solutions if 1-3k ≥ 0, that is, if k ≤ 1/3.
vipin2004:
By the way it's Alana here
Hmm. Should change it.
Are u asking
???
no. thinking aloud
No u should surely not change it
Are u really 28??
Yeh, I know... bit past high school. Just like helping out.
Hmmmm
Great
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