4 y squared + y = 0, y=? Learn Algebra! (Quadratic Equations) 

Why factoring works faster and easier

@@stevenjohnson1143 I know. I got my factoring right but forgot to rewrite my equation. So I got mixed up and went to the QF. My bad. ;)

y*(4y+1)=0 becomes 2 equations since either the y or the (4y+1) can equal 0 and then multiply to become 0. So you are right looking at y=0 would make the equation equal 0 by multiplying, y can also equal -1/4 (4y+1 using y=-1/4 becomes (4*(-1/4) + 1 or -1 + 1) which when multiplied by -1/4 also equals 0. This allows both y= 0 and y= (-1/4) to be solutions to the equation y(4y+1)=0. You can prove both by substituting 0 as y in the original equation (4(0^2)+0=0 which becomes 4*0+0=0 leading to 0+0=0 finally becoming 0=0. And by substituting -1/4 for y in the original equation like this 4(-1/4)^2+(-1/4)=0 negative 1/4 squared becomes positive 1/16 making the equation 4*(1/16)+(-1/4)=0 leading to (4/16)+(-1/4)=0 reducing the first fraction let's the equation become 1/4+(-1/4)=0 and adding both positive and negative 1/4 gives you the 0 in the end. I hope this helps, it is a bit wordy though

If y*(4y+1) = 0 then there are two possibilities: One possibility is that y = 0. The other possibility is that y is NOT 0, but (4y+1) is 0. In the second case, 4y+1 = 0 means that y must be -0.25. So those are the two solutions to the equation in the video: y = 0 and y = -0.25

@@awcampbell2002 I might get back to you when I'm in the mood to read all of that... 🙂

@@fron3107 completely reasonable, gavindeane3670 seems to have given a much shorter explanation that might help so you don't have to wade through mine