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?)
$\displaystyle a-0.5bx^2-.5bx = c-0.5dx^2-.5dx
$
$\displaystyle a-c-0.5bx^2+0.5dx^2-.5bx+0.5dx = 0
$
$\displaystyle -0.5bx^2+0.5dx^2-.5bx+0.5dx+a-c = 0
$
$\displaystyle (-0.5b+0.5d)x^2+(0.5d-0.5b)x+(a-c) = 0
$
Now substitute the bracketed parts into the eqn in post #2.
No program needed, go to the sub-forum titled "Latex Help" read the first couple of posts. All will be clear.
$\displaystyle a - 0.5bx^2 - 0.5bx = c - 0.5dx^2 - 0.5dx$
Yes, try to move everything onto one side.
$\displaystyle a - c - 0.5bx^2 + 0.5dx^2 - 0.5bx + 0.5dx = 0$
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.
$\displaystyle 0.5dx^2 - 0.5bx^2 + 0.5dx - 0.5bx + a - c = 0$
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:
$\displaystyle dx^2 - bx^2 + dx - bx + 2a - 2c = 0$
Now, factor out the x^{2} and the x from the terms that have them:
$\displaystyle (d - b)x^2 + (d - b)x + (2a - 2c) = 0$
This is where it gets ugly. If you want to use the quadratic formula
$\displaystyle x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$
(I'll use capital letters here)
...plug in (d - b) for A, (d - b) for B, and (2a - 2c) for C:
$\displaystyle x = \frac{-(d - b) \pm \sqrt{(d - b)^2 - 4(2a - 2c)(d - b)}}{2(2a - 2c)}$
EDIT: Too slow!
both sides are equal, so "twice both sides" are equal
$\displaystyle 2a-bx^2-bx=2c-dx^2-dx$
both sides are equal, subtracting them gives zero...
$\displaystyle 2a-2c-bx^2--dx^2-bx--dx=0$
$\displaystyle 2(a-c)+(d-b)x^2+(d-b)x=0$
divide both sides by (d-b) if "d" is not equal to "b"
$\displaystyle x^2+x+2\frac{a-c}{d-b}=0$
$\displaystyle x=\frac{-1\pm\sqrt{1-8\frac{a-c}{d-b}}}{2}$
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?
I guess if you take everything but the last line from Archie Meade's post you could complete the square, like this:
$\displaystyle \begin{aligned}
x^2 + x + 2\frac{a-c}{d-b} &= 0 \\
x^2 + x &= -2\frac{a-c}{d-b} \\
x^2 + x + \frac{1}{4} &= -2\frac{a-c}{d-b} + \frac{1}{4} \\
\left( x + \frac{1}{2} \right)^2 &= \frac{-2\frac{a-c}{d-b}}{1} + \frac{1}{4} \\
\left( x + \frac{1}{2} \right)^2 &= \frac{-8\frac{a-c}{d-b} +1}{4} \\
x + \frac{1}{2} &= \pm \sqrt{\frac{1 - 8\frac{a-c}{d-b}}{4}} \\
x &= -\frac{1}{2} \pm \frac {\sqrt{1 - 8\frac{a-c}{d-b}}}{2} \\
x &= \frac {-1 \pm \sqrt{1 - 8\frac{a-c}{d-b}}}{2} \\
\end{aligned}$
I skipped a step here and there, but hopefully you'll get the gist.