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SW-User
[b]2y^2 = 10y + 7
[/b]
This is a quadratic equation so you want to start by getting all the terms to one side so you have zero on the other...
[b]subtract 10y + 7 from both sides
2y^2 - 10y - 7 = 0[/b]
You can't factor this conveniently so you can either stick it into the quadratic formula or solve by completing the square...
[big]Completing the Square![/big]
First you want the coefficient of y^2 to be one...
[b]Divide both sides by 2
y^2 - 5y - 7/2 = 0[/b]
Now you want take half of the coefficient of y then square it and add and subtract it from the left side...
[b] Add and subtract (b/2)^2 = (5/2)^2 =25/4 from the left hand side
y^2 - 5y + (5/2)^2 - (5/2)^2 - 7/2 = 0
y^2 - 5y + (5/2)^2 - 25/4 - 7/2 = 0
y^2 - 5y + (5/2)^2 - 39/4 = 0[/b]
Now you can factor the first three term!!!!
[b](y - 5/2)^2 - 39/4 = 0[/b]
Now you want the squared thing to be on one side and everything else to the other...
[b]Add 39/4 to both sides
(y - 5/2)^2 = 39/4[/b]
Now you want to get rid of the square!
[b]take the square root of both sides
y - 5/2 = ±sqrt(39)/2[/b]
Now finally isolate y
[b]Add 5/2 to both sides
y = 5/2 ± sqrt(39)/2 [/b]
DONE!!!
[/b]
This is a quadratic equation so you want to start by getting all the terms to one side so you have zero on the other...
[b]subtract 10y + 7 from both sides
2y^2 - 10y - 7 = 0[/b]
You can't factor this conveniently so you can either stick it into the quadratic formula or solve by completing the square...
[big]Completing the Square![/big]
First you want the coefficient of y^2 to be one...
[b]Divide both sides by 2
y^2 - 5y - 7/2 = 0[/b]
Now you want take half of the coefficient of y then square it and add and subtract it from the left side...
[b] Add and subtract (b/2)^2 = (5/2)^2 =25/4 from the left hand side
y^2 - 5y + (5/2)^2 - (5/2)^2 - 7/2 = 0
y^2 - 5y + (5/2)^2 - 25/4 - 7/2 = 0
y^2 - 5y + (5/2)^2 - 39/4 = 0[/b]
Now you can factor the first three term!!!!
[b](y - 5/2)^2 - 39/4 = 0[/b]
Now you want the squared thing to be on one side and everything else to the other...
[b]Add 39/4 to both sides
(y - 5/2)^2 = 39/4[/b]
Now you want to get rid of the square!
[b]take the square root of both sides
y - 5/2 = ±sqrt(39)/2[/b]
Now finally isolate y
[b]Add 5/2 to both sides
y = 5/2 ± sqrt(39)/2 [/b]
DONE!!!
Madame · F
@SW-User That's what I got like 18 mins ago!! I posted the answer in the first comment.
ImRileyTheDog · 22-25, F
@SW-User I STILL DON'T UNDERSTAND WHY YOU JUST THROW THE = SIGN TO THE END?!😫
Madame · F
@ImRileyTheDog What do you mean?
ImRileyTheDog · 22-25, F
@Madame Ok so the problem is 2y^2=10y -7 but you take the = sign between the 2 and 10 and put it behind the 7 then = it to 0. Why? I don't recall doing that for some reason.
SW-User
@ImRileyTheDog subtract 10y + 7 from both sides
Madame · F
@ImRileyTheDog You're trying to isolate the 'y' variable from the other stuff. In doing so, it's going to shift the non 'y' stuff to one side.
ImRileyTheDog · 22-25, F
@SW-User OK YESS! I RECALL NOW! Geez I am still in high school and I couldn't recall that for some reason. Thank you! I thought I was going insane.