3y^2+5Y+2 does anybody know how to factor this one?>.
Factorise using quadratic fomulas : http://www.analyzemath.com/Equations/proof_quadratic.html Ans should be y=-1 and y=-2/3
I'm guessing y and Y is the same variable. Your expression is in the form \[ay^2+by+c\] You can apply the quadratic equation: \[{-b \pm \sqrt{b^2-4ac}} \over 2a\]
sorry FOLKS!!but I'm not that satisfied with your answers..tnk u for hitting me though....
do what phoenix said it is a simple quadratic.
\[3y^2+5y+2\] a=3 b=5 c=2 Plug those numbers in the quadratic equation and you get \[y={{-5\pm \sqrt{25-24}} \over 6}={{-5 \pm \sqrt{1}} \over 6}\] \[\sqrt{1}=1\] Therefore, \[y={{-5+1} \over 6}={-4 \over 6}=-{2 \over 3}\] and \[y={{-5-1} \over 6} = {-6 \over 6} = -1\]
@PhoenixFire did u use the method of decomposition in solving the problem>?
No I used the Quadratic Formula which produces two results. Because the forma is + or - the square root. So you have to apply the formula twice, once using + and once use -.
Please have a look here to learn what he did http://www.teacherschoice.com.au/maths_library/algebra/alg_6.htm
aaahh..alright i got you now...
is this right?..5+-/(5)^2-4(3)(2)/2(3)
-5 in the beginning. but yes that's correct.
you forgot square root
square root of what?
No the square root is there... it's just noted by a /
square root of (b^2-4ac) term.
it's actually in the eqUATION...TAKE A LOOK AGAIN...AND SEE IF THE SQUARE ROOT IS MISSING..
SO FINAL ANSWER WOULD BE 2 AND 1
IS MY ANSWER RIGHT?
No. Look at my answer.
WHAT?!!?!? :_(
SO WHAT DO U THINK IS THE RIGHT ANSWER..?
Look above. y=-1 and y=-2/3
AAAH..OKAY.. IT'S JUST THAT YOUR EQUATION IS QUIET CONFUSING..
TNX A LOT...BUT CAN YOU DO ME A FAVOR?..CAN YOU FACTOR THAT EQUATION, FOR ME TO SEE HOW IT GOES///
It's the quadratic equation. It may be confusing, but it's useful to learn and use correctly. Also, can you please drop the capslock.
sorry about..anyway thnx u so much i learned a lot from you today,, ..
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