9y^2+2=9y
Have you considered putting it in standard form so it looks like a Quadratic Equation - the sort of thing you have solved already, many times?
idk how to do that.
add -9y to both sides
thats how it was wrote
You don't know how to put it in standard form? Can you do as phi suggested?
yes
Okay, show the standard form.
-9y+9y^2+2=0?
yes, but you should write it in descending exponents. (the -9y is the same as -9y^1 and it comes after the 9y^2
Descending exponents. Try to make the first coefficient positive.
so 9y^2-9y+2=0
yes. this one can be factored, but it's tricky
how do you do that?
if we multiply the 9 from y^2 times the (last) 2 we get 18 we now list all ways of multiplying 2 numbers to get 18: 1,18 2,9 3,6
the + on the +2 means the roots are the same sign the - on the -9y means the "bigger" root is negative. together , both of these rules means both roots are negative. we look at our list of pairs, making both numbers negative, and add them up -1-18= -19 -2-9 = -11 -3 - 6 = -9 look for the pair that give a -9 (to match the -9y term)
we see -3 and -6 are the pair. the 2 must divide into one of these numbers (the -6) to get -3, -2, 3 this means (how to explain!?) (3y -1)(3y-2)=0
is that the answer? cause this is still so confusing.
you have two terms that could be 0 (and make the equation true) can you solve them? 3y-1=0 and 3y-2= 0
-3 and -1.5
Try that first one again.
3
3y - 1 = 0 add 1 3y = 1 Divide by 3 y = 1/3 You seem to be guessing.
i was dividing it by 1. im not thinking im sorry
It's okay. Take a deep breath. :-) Now check. 9y^2+2=9y Substitute those values and make sure they work.
9(1/3)^2+2=9(1/3) 1+2=3 3=3
One down. How about the other one?
what am i plugging in?
You are not "plugging in" anything. Substitute the value y = -1.5, the other one you determined by factoring.
9(-1.5)^2+2=9(-1.5) is that how i set it up?
That's it. Why would it be different from y = 1/3? Substitute and go!
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