#I dont understand shi--
144 messages · Page 1 of 1 (latest)
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ðŸ˜
now how did this happen
Put $ at the front too
$x^2 - 1$ can be factored as $(x-1)$$(x+1)$
CaliforniaJumpingBean
Do you see that
Wdym how
So the first term is x^2, the middle term is 0x, and the last term is -1
So the middle is canceled
Therefore we have factors that are the same, except for the opposite sign
And since the last term is -1
We know the factors are -1 and 1
Idk
Do you get that?
We can also factor the bottom similarly
how
how is x^2 and -1 same
No
sorry for my low iq actually i am not a math student
-1 and 1
That makes the (x-1) and the (x+1)
So
Basically
We look at the first term for x
It is a singular x
yeah
So the only choices for the coefficients are 1 and 1 right
yeah
aight
So we have no middle term right
yeah
And the last term is -1
So there is only 1 combination of a and b that works
-1 and 1
honestly factoring the top part doesn't really matter
why we use two signs - and + why not negative or positive both sides
it's the bottom part that you should factor
difference of squares in general, $x^2-a^2=(x-a)(x+a)$
Anonymous
i cant even do the basic how will i factor that denominator T_T
just try and distribute and you can convince yourself this always holds true
Do you know the quadratic formula
Or grouping
I suggest you watch some sort of video and learn that first. It's hard to do these problems with that skill.
but the idea for factor $x^2-7x+12$ is you find two numbers whose product is 12 and sum is -7
Anonymous
x = -b+- root b^2*4ac/2a
Can you try this
wait wait this is way too hard for me
let me do it
should i use quadratic formula to solve this
try and think about it first
no
i mean you can but this is much faster
Good luck with this problem
I'll leave it up to you to help
understood
I think 2 of us isn't helpful
(x-3)(x+4)
ya sorry i kinda just interrupted from nowhere though ðŸ˜
whats -3+4
1
well that's not -7
perfect
less go
here's how I'd approach this problem:
$\frac{x^2-1}{x^2-7x+12} \ge 1$ \
$\frac{x^2-1}{x^2-7x+12}-\frac{x^2-7x+12}{x^2-7x+12}\ge 0$ \
$\frac{7x-13}{(x-4)(x-3)}\ge 0$
huh
Anonymous
I subtracted 1 from both sides and got a common denominator so that i can add the fractions
so did you cross multiply denominator with 1 to get it common
it's not really cross multiplying
you just want the bottom part to be the same
i didnt understood after subtracting 1 from both sides , how did you got same denominator
i multiplied the top and bottom by x^2-7x+12
well notice, what is $\frac{x^2-7x+12}{x^2-7x+12}$ equal to? the same number divided by itself
Anonymous
yeah
well, what is it?
1 ?
yes but you can't add the fractions if the bottom term isn't the same
if you just had 1 you can't add that
should i use the wavy curve method to further solve this ?
i do not know what that is
its like a signchart to find out the intervals
i have my own way of doing it which is pretty fast but i'm not sure if it'll help you or confuse you more
if you want to, sure
i dont want to waste time on long methods
but please elaborate while explaing cuz i am just dumb
dash lines means negative, bold lines means positive
if you are familiar with the quadratic and linear functions,
(x-4)(x-3): you know that it's negative in between the roots and positive everywhere else
(7x-13): the line is positive for anything above its root (13/7) and negative everywhere below it
multiply the signs of both regions, and you get your desired result
it's like the same process of how people normally do it, except if you understand the functions well enough you don't need to plug anything into your calculator
bring one to this side, make it into a single fraction, create factors, check od powered factors for their zeroes and plot it on number line using wavy curve
never understood why people just restate the already answered
well i didnt knew that one
thanks for explaining it was really helpfull ^^
i solved it ty