How do I solve for x here? I know I need to move things over to one side but I don't know how to handle the x²:
a-.5bx²-.5bx = c -.5dx² -.5dx
TIA.
Thanks, pickslides. Unfortunately I need more information. This is for work and I unfortunately haven't had algebra in decades. I recognize the reference to using the quadratic equation and am guessing:
on left side:
a= -.5b
b= -.5b
c= 1
on the right side:
a= -.5d
b= -.5d
c= 1
Would I then move one quadratic equation over to the other side by subtracting it from each so that everything equals '0'?
or do I move everything over first and then plug them into a quadratic equation, ie:
a-.5bx²-.5bx = c -.5dx² -.5dx
a-c-.5bx²-.5bx +.5dx² +.5dx = 0 (/edit: +.5dx² +.5dx)
If this is the case, I'm confused what a,b,c would be with so many variables.
As you can see, it's been a long time.
(Btw, what program do you use to create your equations and to display as an image?)
Yes, try to move everything onto one side.
It's helpful to rearrange the terms so that the x's are in decreasing order -- first the x^{2} terms, then the x terms, then the constant terms.
You'll notice that I also put the terms with "d" in front of the terms with "b". I did that because I like putting positive terms first.
Before we go on I am going to multiply both sides by 2 to get rid of the decimals:
Now, factor out the x^{2} and the x from the terms that have them:
This is where it gets ugly. If you want to use the quadratic formula
(I'll use capital letters here)
...plug in (d - b) for A, (d - b) for B, and (2a - 2c) for C:
EDIT: Too slow!
Thanks so much everyone. It's really helpful seeing the steps taken. Each of them are very understandable (and actually pretty cool) altho definitely not something I could have done at this rusty stage without your help.
I'm curious, what would another solution from here look like without using the quadratic formula? Does it matter that X will always be >0?