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

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[@1234larry1]

Anytime a quadratic doesn't have a constant (ordinary number) on the end, It can be solved by factoring out the variable. This gives you one variable=0 and one linear equation=0, so one solution is always 0 and the other is the solution you get when you solve the linear equation for the variable.

[@mylittlepitbull3143]

Hey buddy really appreciate the review on all the math. I need something a little more difficult, though.

[@Kleermaker1000]

Two solutions here: y(4y + 1) = 0, so y = 0 and y = - 1/4.

[@louf7178]

This is a good video. I'm not sure the blur of words is comprehendible to the beginner and is why this review is valuable.

[@aryusure1943]

I got it and would never have done it a few months ago when I began watching your videos. I was on my way with factoring but I made it interesting with the QF. ;)

[@stevenjohnson1143]

Why factoring works faster and easier

[@aryusure1943]

@@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. ;)

[@BrandurDavidsen-wd2lp]

-1 +/- (sqr 1^2 - 4 x 4 x0)/2 x 4 (-1+/- sqr 1)/8 y = 0/ 8 y = -2/8= - 0,25

[@stevenjohnson1143]

Factoring y out gives us y(4y+1)=0 setting both factors equal to 0 you get y=0 y=-1/4

[@raya.pawley3563]

Thank you

[@erembald2927]

solved it numerically. Didn't occur to me to find the zero product, I'll remember next time. good one.

[@russelllomando8460]

got both 0 & -1/4 factor the Y and both =0 thanks for the fun

[@inkheart151]

But why isn’t the inside y also zero and multiplied by 4?

[@chrisdissanayake6979]

Y = 0. or Y = - 1/4

[@stevewthespider]

( 0 , -0.25 )

[@3060Gaming-t2i]

I get it👍

[@maxinemcafee4893]

Y=0 , y= -1/4

[@franklinbrown5625]

Y(4y+1)=y=0 or 4y+1=0. So y=0 or 4y=-1, y =-1/4. Prove: 4(-1/4)^2+(-1)=1/4+(-1/4)=1/4-1/4=0

[@dailydriven031]

Read a book of music business. It can help. Put together your brain. For this equations.

[@MichaelLong-k1u]

2

[@fron3107]

If y*(4y + 1) = 0 and y = 0, than 0*(4*0 + 1) = 0. Anything times 0 is 0 - so how can you state 4*0 + 1, which is 1, is 0??

[@awcampbell2002]

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

[@gavindeane3670]

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

[@fron3107]

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

[@awcampbell2002]

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

[@larrnew]

0

[@josephlaura7387]

y=0, y = -1/4