3b^2 + b-2 =0 please help solve
\(\bf{3b^2 + b - 2= 0}\) You can solve this by splitting the middle term , here it is "b" . I can write b as "3b - 2b" ? , right?
okay
Now , I can write the equation as : \(\color{blue}{\bf{3b^2 + 3b - 2b - 2 = 0 }}\) Or : \(\color{purple}{\bf{(3b^2 + 3b)}} \color{orange}{\bf{-2b - 2}} = 0 \)
Now, \(3b^2 + 3b\) can be written as : \(3b(b + 1)\) , right?
oka I am following you because I am not sure how to work
Also, -2b-2 can also be written as : -2(b+1) So, 3b(b+1) - 2(b+1) = 0 (3b-2)(b+1) = 0
so the answer to the question will be the last thing you wrote correct?
No, (3b-2) ( b+1) = 0 So, either 3b - 2 = 0 that is b = 2/3 or b + 1 = 0 , b = -1 So either b = 2/3 or b = -1
so how do I put all of this together?
are you there?
Oh sorry, I was away for some time. What do you mean by "SO how do i put all of this together?"
I need to know how to put all of the problem that you have work together so that I can see each step to better understand it
Okay I will write it for you : You're given with : \(3b^2 + b - 2 = 0 \) "b" can also be written as 3b - 2b So, the equation becomes : \(3b^2 + 3b - 2b - 2= 0\) or : \(3b(b+1) - 2(b+1) =0\) \((3b-2)(b+1) = 0 \) Either : 3b - 2 = 0 or b + 1 0 3b - 2 = 0 , b = 2/3 or b + 1 = 0 , b= -1 So either b = 2/3 or b = -1
Thank you so much
I have another problem is 4x^(3.5) a polynomial and what is the domain range of it I am not algebra smart so I need all the help I can get
You're welcome :) \(\Huge{\color{purple}{\textbf{W}} \color{orange}{\cal{E}} \color{green}{\mathbb{L}} \color{blue}{\mathsf{C}} \color{maroon}{\rm{O}} \color{red}{\tt{M}} \color{gold}{\tt{E}} \space \color{orchid}{\mathbf{T}} \color{Navy}{\mathsf{O}} \space \color{OrangeRed}{\boldsymbol{O}} \color{Olive}{\mathbf{P}} \color{Lime}{\textbf{E}} \color{DarkOrchid}{\mathsf{N}} \color{Tan}{\mathtt{S}} \color{magenta}{\mathbb{T}} \color{goldenrod}{\mathsf{U}} \color{ForestGreen}{\textbf{D}} \color{Salmon}{\mathsf{Y}} \ddot \smile }\) If you liked my response, please click on Best response next to my replies.
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