Use the quadratic formula to find the zeros of the function. Round to tenths if necessary. y = 17x2 - 21 So since its not in ax^2 + bx + c wouldnt it be no real solution? A. {-1.1, 1.1} B. {1.1} C. {-4.1, 4.6} D. no real solution
why isn't it \[ax^2 + bx + c\] ? b is simply zero...
oh i seeee wow my dad told me otherwise :P
although it should say \[0 = 17 x^2 - 21\] (the zero _is_ important) hope that was just a typo on your side? Got the result yet?
yea it was :P lol cuz i accidentally copied and pasted but not all of it :P lo l
haha
the answer is A - not B... (cause you cant tell if a number was positive or negative after squaring it... hence the +/- in front of the sqrt in your formula...)
ohhhh i seeee :) thx alex lol
ur not rly the same as u used to be :P
what's that supposed to mean? :P
ur acting like my nerd friend not my friend friend lol :)
( i think it's the fame, btw ^^)
o.O that's not nice! ( i think ^^) just cause you lost a fan? i'll fan you again, when you try harder.. and not run off with some other boy :'-(
@MuH4ha : Btw you missed our study session. I dont know if it was on purpose or not, but the lesson i was doing today was rly complicated and i needed ur help!! :P
@Dolly: I'm sorry - it won't happen again... I promise ;) Had a nice Zombie-Jesus Day? Hope to hear from you soon! xo
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