How do I solve 8x^2+2x=12x?

 

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ekmccall12309 (author) 7 years ago
 actually..i got my mom to help me (she's a math teacher) and after you move 12x over and set it equal to zero you get 8x^2 - 10x = 0 then you pull out 2x and get 2x(4x - 5) = 0 then you set both parts equal to zero and you get x=0 and x=4/5. just to clarify for everyone trying to solve this for me. thanks anyways for guys!! i really appreciate it. :)
ekmccall12309 (author) 7 years ago
 also in reply to on of MahvishnuMan's post..0 can be the answer for any variable if the equation is set equal to zero and if there are no numbers without variables.  and also as a "modern day student" myself, we DO do a fair amount of graphing. and as a fix to my last post..i meant "thanks anyways guys!!"
The only possible outcome is x=0.<br /> <br /> Normally, your first move trying to solve the equation is to move all the variables to one side of the equation and leave nothing but constants on the other. The problem with this equation is that there are no constants that you can use to define the x's. Therefore, the only possibility that makes sense is x=0; it's the only number you can plug into x that works.<br />
Not true.  8x2 = 10x solves for the non-zero solution x=1.25.
Yeah, you're right.  I didn't think to graph it.
That's not graphing. It's simple algebra.
8x2 + 2x = 12x
8x2 = 10x
8x * x= 10 * x (This step isn't necessarily, it's just illustrating the previous step)
8x = 10
x = 5/4, or 1.25
Yeah, I know.  I read the solution that was provided.  However, since zero also solves the equation, I was at least partly correct.  If I would've graphed it, I would've seen that the plot crosses the x axis at two points.

I never claimed algebra was my strong suit.  It's been awhile since I used this stuff.
Actually, as a modern day student, we do very little with graphing. Algebraic solutions are much faster. Which is why I might have posted this, because of "I didn't think to graph it."

Sorry for the inconvenience.
I'm stupid.  You're the king of modern algebra.
Divide everything by x and you'll get a simpler solution, which coincides with the other zero reached by A_of_s_t.
The graphical solution, as before described:
8x squared.bmp
Hey a graphical answer!  Nice one!  If I may ask:  What's the application that made that pretty picture?
Hmmm... GraphCalc
http://www.graphcalc.com/what_is_graphcalc.shtml
http://en.wikipedia.org/wiki/GraphCalc

Looks interesting.  Thanks for the reply.
A_of_s_t7 years ago
8x2 + 2x - 12x = 0<br /> 8x2 -10x = 0<br /> x(8x - 10) = 0<br /> <br /> x = 0<br /> <br /> 8x-10 = 0<br /> x = 10/8 = 1.25<br /> <br /> <strong>x = 0, 1.25</strong><br /> <br /> Check the solutions:<br /> <br /> 8(0)2 + 2(0) = 12(0)<br /> <strong>0 = 0 </strong><br /> <br /> 8(1.25)2 + 2(1.25) = 12(1.25)<br /> 12.5 + 2.5 = 15<br /> <strong>15 = 15</strong><br /> <br /> For a graphical solution, graph the parabola 8x2 + 2x and the line 12x, their intersections are the solutions to the problem. <br /> <br /> The degree of the equation -- the largest power of x -- states the number of zeros the equation will have. If the largest power of x is 2, then there are two solutions to the equation.
Well said.
Incidentally, this editor requires a caret before and after the superscript text...
Unfortunately, the "caret" for superscript is a temporary hold-over from the old text-based markup system, and will disappear soon, once the I'bles servers have completed running through and converting all of the old comments in the database to the new format.

At that point, the "extended markup" (sub- and superscript, big and small text, etc.) will be available only to paid account.  A group of us users are advocating with Eric to change that policy (so that pre-existing markup is available to all users), but no decision has been made yet.
Heh. Is that why I keep seeing </br> everywhere suddenly?
Rewrite the equation as : some polynomial in x = 0
Like this:  8x2 -10x +0 =0

Now factor it:
8x2 -10x +0 =0
8x(x -10/8) =0
x(x-5/4) =0

And you can see the answer, x=0 or x=5/4, from inspection.

Check your work using Octave or Matlab. Write the poly as a vector of its coefficients, then feet that to the roots() function.
octave.exe:1> roots([8,-10,0])
ans =

   1.25000
   0.00000

Oops...  I meant feed that, not feet that.
XD
kelseymh7 years ago
Since there's no constant term, you don't have to do any factoring.  Divide through by "x" and solve by inspection.
origamiwolf7 years ago
x = 0.<br />
Z..7 years ago
Holy dooley!!

Well done logangina for even attempting it!
8x2 + 2x =12x<br /> 8x times 8x+2x =12x<br /> 64x+2x=12x<br /> 66x=12x<br /> which makes no sense<br /> <br /> i believe you need a number that is not an "x" number to solve this<br />