Question

what is the solution set of the equation absolute value 4a+6 close absolute equation -4a = -10

Answer 1

|4a+6|-4a=-10

Answer 2

yes do you make it into 2 separate equations?

Answer 3

1. Add 4a to both sides
2. Square both sides
3. Put everything to the left
4. Solve the quadratic equation

Answer 4

Don't need to separate :)

Answer 5

okay ty again

Answer 6

no problem :)

Answer 7

Actually, do you prefer us guiding you to the answer or just give you the steps? :)

Answer 8

guiding the answer

Answer 9

i see

Answer 10

beacause i added 4a to both side but what do you mean by square it

Answer 11

oh to get rid of the absolute value

Answer 12

So it's not a quadratic equation... hmm...

Answer 13

i have to find the solution set

Answer 14

But a=0.5 doesn't seem correct :S

Answer 15

I don't think so either

Answer 16

@ocelot40 yep...

Answer 17

http://www.wolframalpha.com/input/?t=crmtb01&f=ob&i=%7C4a%2B6%7C-4a%3D-10

Answer 18

No solution LOL

Answer 19

I just got more confused

Answer 20

ok w8

Answer 21

ok

Answer 22

Im not sober but ill try to explain :D just a sec

Answer 23

lmfao ok

Answer 24

|x| = x for x>0 and -x for x<0 do you understand it?

Answer 25

This is how absolute value of x is defined

Answer 26

this is the problem can you solve it

Answer 27

If you get it, tell me to proceed

Answer 28

Yes I can :D
Do you want me just to give you the solution or make you understand?

Answer 29

can you go step by step

Answer 30

So, pluging in a=0.5 to |4a+6|-4a=-10, you'll find that the equality isn't valid. Therefore a=0.5 is an extraneous solution.

Answer 31

Therefore there exists no solution what-so-ever at all

Answer 32

yes i understand its no solution but i need to show my work

Answer 33

I gave you all the steps lol

Answer 34

its to confusing to read

Answer 35

Ok, i'll give it a try for you

Answer 36

my teacher is going to ask me where i got .5 from

Answer 37

you subtract 4a from both sides then you get \[\left| 4a+6 \right|\] = -10 +4a

Answer 38

\[\large\begin{array}{rcl}|4a+6|-4a&=&-10\\|4a+6|&=&4a-10\\(4a+6)^2&=&(4a-10)^2\\16a^2+48a+36&=&16a^2-80a+100\\128a&=&64\\a&=&0.5\end{array}\]
Check:
\[\text{When }a=0.5,\]\[\begin{array}{rcl}L.H.S.&=&|4(0.5)+6|-4(0.5)\\&=&|2+6|-2\\&=&8-2\\&=&6\\R.H.S.&=&-10\end{array}\]\[\therefore\text{a=0.5 is extraneous.}\]\[\therefore\text{There exists no solution of any kind what-so-ever at all.}\]

Answer 39

I missed the line \(\because L.H.S.\ne R.H.S.\)

Answer 40

Above the first \(\therefore\) sign.

Answer 41

thank you so much

Answer 42

no problem :)

Answer 43

Or you can, as @ocelot40 said, split it into two cases.

Answer 44

Like every absolute value, we have to check for what x's is positive and for what x's is negative

Our absolute value is |4a+6| so we take 4a+6=0 => a=-3/2 from this we get that for values of a less than -3/2 4a+6 is negative
and for values of more or equal than -3/2 4a+6 is greater or equal than zero.

We are going to take both cases: 1) a<-3/2 and b) a>-3/2

a) a>-3/2 ---------------- |4a+6|=4a+6
our equation becomes 4a+6-4a=-10
6=-10 So we see it's impossible for a>-3/2

b) a<-3/2 ----------------- |4a+6= -4a-6
our equation becomes -4a-6-4a=-10
a=2 BUT we said in the beggining that a<-3/2 so we cant take a=2 for an answer

So either case the solution is impossible!

Answer 45

@ocelot40 Why don't you say a) 4a+6>0 and b) 4a+6<0 just suggesting no offense
I appreciate your effort :D

Answer 46

thank you i already finished it

Answer 47

@kc_kennylau It's the same thing but saying a>-3/2 is more simplified

Answer 48

but more comprehensible :)

Answer 49

i have another question if you don't mind

Answer 50

@maddiec sorry I need to have breakfast now (GMT+8) so see ya :D In the meantime @ocelot40 will help you :)

Answer 51

\[\Huge\frac{\sqrt{108x^5y^8}}{\sqrt{6xy^5}}=\frac{\sqrt{108}}{\sqrt{6}}\times\frac{\sqrt{x^5}}{\sqrt{x}}\times\frac{\sqrt{y^8}}{\sqrt{y^5}}\]

Answer 52

But you have at least to try that yourself... Where're you stuck at? :)

Answer 53

the variables throw me off

Answer 54

Don't be scared by the variables :)

Answer 55

so the first set i put it all under one radical?

Answer 56

yep :)

Answer 57

the after i have to see what goes into 108

Answer 58

nvm, ocelot's helping you already :)