#i need a little help with this
feral falconBOT
sharp valley
NO
no*
null factor law only works when mn = 0
when you have two numbers that multiply to give zero
at least one of them has to be zero
when mn equals anything else
theres an infinite amount of pairs that can satisfy the equation
loud iron
Sorry if i bothered you but like i can't do anything else if the equation is equal to number? The only thing i can do is to make it equal to zero and then solve it?
loud iron
Can i use a quadratic formula if it equals a number?
sharp valley
well if it equals a number
say your equation is
2x^2 + 3x + 5 = 7
you have to bring the 7 over
because the form is ax^2 + bx + c = 0
trim stump
It's true that if any 2 numbers in a multiplication is 0 then the whole product is 0, but it's not true that if any 2 numbers is -1 then the product is -1
loud iron
Got it thanks mate
trim stump
You probably already know but your equation can be written as (x+1)^2=0
At which point you can use null factor magic
sharp valley
^
balmy fossil
irrelevant to what the other guy is saying but one of your answers in this case are incorrect since if x=-3 then x*(x+2) = 3
loud iron
You probably already know but your equation can be written as (x+1)^2=0
Yeah i
Did it
So at the end i can only do that when the equation is equal to 0
balmy fossil
no the function x^2 +2x+1 is not linear
what you have is the graph of the solution
which there is just a single number
-1
devout sage
no the function x^2 +2x+1 is not linear
no cope
loud iron
Isn't x^2 is like this (U)
balmy fossil
Isn't x^2 is like this (U)
indeed, it describes a parabola
loud iron
indeed, it describes a parabola
Yeah
devout sage
no the function x^2 +2x+1 is not linear
y = x^2 + 2x + 1
its a joke
man
and my statment is right this is linear
you cope
trim stump
dang
topaz wren
Simplify [(x⁴y‐¹)²/x⁵y‐³]² × (x‐²y⁶)/2(xy³)‐² 26 y⅓/x½ ÷ (x½y⅔/x³y⁴)
devout sage
÷ 💀
gaunt archBOT
TimK
earnest sluice
quadratic equation
loud iron
.solved