Question

solve for y: y^2-81=0
this is solving factorable quadratic equations
medal awarded

Answer 1

this is the difference of 2 squares
a^2 - b^2 = (a + b)(a - b) = 0

now substitute y^2 for a^2
and 81 for b^2

what do you get?

Answer 2

y^2-81=(y+81)(y-81)

Answer 3

not quite
b^2 = 81
so b = 9

follow?

Answer 4

y^2 = 81
so y = 9

Answer 5

yes okay so y^2-81=(y+9)(y-9)

Answer 6

how is y^2=81

Answer 7

compare your equation with the general form:

a^2 + b^2

y^2 + 81

Answer 8

ohh '-' not '+'

Answer 9

so shouldnt it be -81 and not 81

Answer 10

??

Answer 11

@terenzreignz please help im all turned around now gah

Answer 12

I'm terrible at teaching how to factor... as much as it pains me to say it... you need a second (third?) opinion :)

Answer 13

although, with this one, what I can say to you is that if you a difference of two squares..
\[\Large a^2 - b^2\]
then it factors into the sum and difference of their square roots..
\[\Large a^2 - b^2 = (a+b)(a-b)\]

Answer 14

With that in mind....
\[\LARGE y ^2 - 81 = \color{blue}{y^2 - 9^2}\]

Get to it, champ :)

Answer 15

Wait a minute, you already finished this :/

Answer 16

so it is -81 and -9

Answer 17

?

Answer 18

No... lol
(y + 9)(y - 9) = 0

So either
y + 9 = 0
or
y - 9 = 0

You have to consider them both :)

Answer 19

9 and -9

Answer 20

That's right :D

Answer 21

i dont get it

Answer 22

lets do another ill post it

Answer 23

Okay... look at this way...
if the product of two numbers is 0...

ab = 0

then at least one of them must have been zero, right?
So, either
a = 0
or
b = 0


It's not that different with polynomials...
(x+9)(x-9) = 0
Then either
x+9 = 0
or
x-9 = 0

It's that simple ;)