@jagged crest
#|x+3|+|1-2x|=5
glossy heath
Try finding the extent of each
And see where they overlap
This can be done using drawings at different heights
jagged crest
im not sure what you mean
by the extent of each
i think if thats what your asking
proud barn
(|x+3|+|1-2x|)² = 25
(x+3)²+(1-2x)²+2|x+3|•|1-2x| = 25
x²+6x+9+1-4x+4x²+2|x+3|•|1-2x| = 25
5x²+2x+10+2|x+3|•|1-2x| = 25
2|x+3|•|1-2x| = 15-2x-5x²
4(x+3)²(1-2x)² = (15-2x-5x²)²
4(x²+6x+9)(1-4x+4x²)
= 225+4x²+25x⁴-60x-150x²+20x³
4(x²-4x³+4x⁴+6x-24x²+24x³+9-36x+36x²)
= 25x⁴+20x³-146x²-60x+225
4(4x⁴+20x³+13x²-30x+9)
= 25x⁴+20x³-146x²-60x+225
16x⁴+80x³+52x²-120x+36
= 25x⁴+20x³-146x²-60x+225
9x⁴-60x³-198x²+60x+189 = 0
3x⁴-20x³-66x²+20x+63 = 0
,w 3x⁴-20x³-66x²+20x+63 = 0
red havenBOT
glossy heath
im not sure what you mean
Oh sorry, I accidentally misread it as x+3>0 and 1-2x<5 or something
It was late
rare dove
(|x+3|+|1-2x|)² = 25
(x+3)²+(1-2x)²+2|x+3|•|1-2x| = 25
x²+6x+9+1-4x+4x²+2|x+3|•|...
at what stage in math do you learn stuff like this?
lofty fog
at what stage in math do you learn stuff like this?
Hmm that's algebra
rare dove
because I have learnt about graphing absolute values, but would not know how to solve a question like this
although I am just learning through Khan Academy courses 
lofty fog
because I have learnt about *graphing* absolute values, but would not know how t...
Basically what he did was to get rid of the absolute values by taking the power of 2
That way, the thing inside will always be positive
rare dove
ahh okay
lofty fog
And then, you can get rid of the absolute value
rare dove
interesting
lofty fog
You can try it out yourself if you want to
rare dove
if you have
|x+3| for example and raise it to the power of two
do you just expand it like any other binomial?
lofty fog
Yup
lofty fog
maybe I'll try it now
There's just one thing that you have to look for
lofty fog
For example, here you get 4 solutions
But only two of them actually satisfy the first equation
That happens because the ranges of the absolute value and stuff
So whenever you get your final solution, check that it actually work
rare dove
jesus this is a lot of expanding 
muted geyser
(|x+3|+|1-2x|)² = 25
(x+3)²+(1-2x)²+2|x+3|•|1-2x| = 25
x²+6x+9+1-4x+4x²+2|x+3|•|...
A much better approach would be to define the LHS as a piecewise function
rare dove
A much better approach would be to define the LHS as a piecewise function
how do you do that?
muted geyser
how do you do that?
uhm wdym
rare dove
I don't get what "defining as a piecewise" function means
sorry for the ping btw, forgot to turn off
could you explain if you have the time?
muted geyser
hmmm
a piecewise function is basically
A function broken into pieces
I don't know how to explain lemme just
rare dove
I know what a piecewise is
muted geyser
Oh
rare dove
I don't understand how you are suggesting he applies it
rare dove
thanks
muted geyser
Define $f(x) = |x+3| + |1-2x| = |x+3| + |2x-1|$
\Then, notice that for $x < -3$, both $x+3$ and $2x-1$ are negative
\Hence, for $x < -3, f(x) = -x-3-2x+1 = -3x-2$
\For $-3 \leq x < \frac{1}{2}$, we have $x+3 \geq 0$ and $2x-1<0$
\Hence, for $-3 \leq x < \frac{1}{2}, f(x) = x+3 - 2x + 1 = -x+4$
\For $x \geq \frac{1}{2}$, we have $x+3 > 0$ and $2x-1 \geq 0$
\Hence, for $x \geq \frac{1}{2}, f(x) = x+3 + 2x - 1 = 3x+2$
\Using these facts, the whole piecewise definition for $f(x)$ becomes $$f(x) = \begin{cases} -3x - 2 & \text{ if } x < -3 \ -x+4 & \text{ if } -3 \leq x < \frac{1}{2} \ 3x+2 & \text{ if } x \geq \frac{1}{2} \end{cases}$$
then just solve for f(x) = 5
@rare dove
red havenBOT
omegusnyan
