How to solve absolute value equations by understanding "distance"

[@bprpmathbasics]

Solving x^2+9=0 kzbin.info/www/bejne/n6S1apivhZV8Zrssi=qui-VDLGuP82pqDv

[@alejandrocolazo1771]

Not hating at all but why do these videos keep popping up in my feed at 5 a.m. just when i'm about to sleep? I now am obligated to watch them.

[@Brid727]

not a good timezone for you when he posts the videos i guess for some people it'll be a good time for him to post the video and bad time for others and whether he changes the video release time or not it's always the case

[@Kero-zc5tc]

He’s in a United States timezone

[@HerrFinsternis]

Why are you checking your feed though?

[@DatBoi_TheGudBIAS]

Why are u *about* to sleep at 5am?

[@Catornado]

7:53 for 3 or more, I would consider every individual range of values. For example, the original equation you have |x-2| + |x-3| = 2: I would consider x

[@fiprandom3783]

I'd do that too, even with 2 absolute values, but as bprp said, he didn't want to do those things on that video, so I don't think that's what he's looking for

[@MrKoteha]

He wanted a nice way of doing it like in the video

[@ItsPungpond98]

Just use a number line, and set the x values that will make the absolute value be 0 as the critical number. And then, see if the numbers in each interval makes the inside negative. If so, remove the absolute values sign, and negate it. Else, just keep it. You'll have only n + 1 cases to solve, rather than 2^n by separating cases by their signs. (n is the number of absolute values)

[@tygrataps]

Prime Newtons has some lovely videos that do just that! ... and now I'm curious if mathtubers get writing space envy...

[@AhoyYoutube]

bro has taught me more factoring than khan academy every has, bprp you are saving me from not continuing my advanced math courses.

[@IITIAN_dost]

From graphing ❤

[@Fperm-t6g]

you can do #2 with contradiction. x>3→x-2+x-3=1/2→2x=11/2→x=11/4

[@narfwhals7843]

D1+D2=C is the set of points equidistant from 1 and 2. That's an ellipse in R². The solutions in R would then just be the points with y=0. The set equidistant from 3 distinct points is a less nice shape, but if they lie on a line no point is equidistant from them in R². Correct?

[@dannyyeung8237]

Instead of just real solutions, can I find the complex solutions to these 3 equations too?

[@JayTemple]

I always checked my solutions. That said, once you got to 2(expression) + 1 = 2, I would have just gone 2(expression) = 1, (expression) = 1/2 and then solved for x.

[@عليعليعلي-ظ9خ6ج]

18-6i/1+9x18=1-2yi/1-yi+2y2 pleas solve this equation this from my teacher

[@enterlessguy]

Well you could try a table of signs, basically we turn the absolute values into piecewise functions for different intervals and solve each equation fir that interval, if the solution fits within the interval then that's a root for our total absolute value equation, all though I admit its not exactly "nice", buy this allows you to solve literally any absolute value statement

[@dannyyeung8237]

By the way |x|=sqrt(Re(x)^2+Im(x)^2)

[@HerrFinsternis]

I think with three absolute values you would only ever get one or zero answers (given there's three seperate numbers there that aren't x) so you can probably just solve it using algebra?

[@nasancak]

Actaully, to get 1 solution, the three numbers should be picked specially; otherwise there would be more than one solution, if any

[@Bruh-bk6yo]

Proud to say I know this kind of graph is just a saddle :)

[@seanwilkinson7431]

So, looking at the problem and your math, couldn't you just combine the AVs as 2x-5 and -2x+5? Does that work consistently for this kind of problem?

[@ronaldking1054]

In between 2 and 3, one is positive, and the other is negative, which results in the lefthand side equaling 1.

[@enterlessguy]

@@seanwilkinson7431 unfortunately it doesn't... I believe that ONLY works if the coefficient of the linear functions inside the absolute value are the same, and even then if you do that you wouldn't be able to realise that y=2 for a certain x inside a certain interval. Its best to break them down into multiple piecewise functions and solve for each one to see if its in the interval or not

[@cubefromblender]

Is there any answer of x×y=0 where x≠0 and y≠0

[@うみうし-c4u]

f(x)=|x-2|+|x-3| =(x-2)+(x-3) if x>=3 or (x-2)-(x-3) if 2

[@dannyyeung8237]

What about for complex number x

[@ronjharedrobles2960]

Woahh

[@seanbastian4614]

For the second equation, is it no solutions or is it no real solutions?

[@leleteri1069]

no solution, if you draw a complex plane and put x somewehre on there. the relation beetwen x, 2, and 3 are like triangle. a side of triangle always smaller than the sum of the other 2 sides. so even in non real value of x, |x-2|+|x-3|>|2-3|

[@paddle_shift]

1-2=1?

[@3leggedfish41]

No no, the absolute value of 1-2=1 As in, in a number line, the distance between one, and two, is one. Since absolute values measure distances between two points; and distance cannot be negative, |2-1| is the same thing as |1-2|. Since either way you measure it both points will still be one unit away from each other