# Thread: Literal Equations

1. ## Literal Equations

Help, please.
I understand the basics of solving literal equations. I can solve basic equations like P = srt.

But I am stumped on these:

x + SQRT(x) = -y
Solve for x

l/w = w/(l – w)
Solve for w

Can anyone help??
Thanks for any assistance anyone can provide!!!

2. Originally Posted by n2hdcycles
Help, please.
I understand the basics of solving literal equations. I can solve basic equations like P = srt.

But I am stumped on these:

x + SQRT(x) = -y
Solve for x

l/w = w/(l – w)
Solve for w

Can anyone help??
Thanks for any assistance anyone can provide!!!
LITERAL?!? You mean linear equations.
Well, these equations are called quadratic equations, meaning the $\displaystyle x$ vaiable is now squared. Have you studied those?
I can solve them for you but if you never studied quadradic equations then it would be useless to show you how to solve them.

3. ## literal (??) equation

Yes I understand the quadratic formula. Any help in guiding me through the solutions would be appreciated. I am really stumped on these two.

It is my understanding that a literal equation is a linear equation that is all letters (no numbers). "Literal" means "related to or being comprised of letters". See the following link:
http://www.purplemath.com/modules/solvelit.htm

Like I said, I can get basic equations, like P = 2l + 2w solved for l.
P = 2l + 2w
p - 2w = 2l
(p - 2w)/2 = l

It is the square root on the first equation and the fraction on the second equation that's throwing me.

Again -- appreciate the assistance!!

4. Do you know the quadratic formula? If you do I can guide you through those two.

5. Yes, I do understand the quadratic formula.
Sorry, after I posted a reply. I realized I neglected to answer that question.
I've spent many hours attempting to work these problems out.... very frustrated......mmmmmmmmmm...and slightly grouchy for a sunday night

6. For the first problem
$\displaystyle x+\sqrt{x}=-y$
You have to do the following trick, let $\displaystyle \sqrt{x}=z$ thus, $\displaystyle x=z^2$. Substitute that back to the original equation:
$\displaystyle z^2+z=-y$
$\displaystyle z^2+z+y=0$
Use quadratic formula,
$\displaystyle z=\frac{-1\pm \sqrt{1-4y}}{2}$
But, $\displaystyle z=\sqrt{x}$
$\displaystyle \sqrt{x}=\frac{-1\pm \sqrt{1-4y}}{2}$
Now square both sides (do one for plus and then for minus):
Remember to use F.O.I.L.

$\displaystyle x=[\frac{-1+\sqrt{1-4y}}{2}]^2=\frac{2-4y+2\sqrt{1-4y}}{4}$=$\displaystyle 1/2-y+\frac{\sqrt{1-4y}}{2}$

Now do the same thing for the minus,

$\displaystyle x=[\frac{-1-\sqrt{1-4y}}{2}]^2=\frac{2-4y-2\sqrt{1-4y}}{4}$=$\displaystyle 1/2-y-\frac{\sqrt{1-4y}}{2}$

I would say if you are starting to learn algebraic equations this equation was very difficult

7. For the second,
$\displaystyle \frac{l}{w}=\frac{w}{l-w}$ for $\displaystyle l$
Cross multiply,
$\displaystyle l(l-w)=w^2$
Open parantheses,
$\displaystyle l^2-wl=w^2$
Organize to quadratic form,
$\displaystyle l^2-wl-w^2=0$
Apply Qudratic formula,
$\displaystyle l=\frac{w\pm \sqrt{w^2+4w^2}}{2}$
Add them up,
$\displaystyle l=\frac{w\pm \sqrt{5w^2}}{2}$
But, $\displaystyle \sqrt{5w^2}=w\sqrt{5}$
Thus,
$\displaystyle l=\frac{w\pm w\sqrt{5}}{2}$
Factor $\displaystyle w$
$\displaystyle l=w[1\pm \frac{\sqrt{5}}{2}]$

It is interesting to note this problem was about the Golden Mean which appears in this solution represented by $\displaystyle \phi$ it is the positive solution for $\displaystyle l$ thus, $\displaystyle \phi=\frac{1+\sqrt{5}}{2}$. It makes many interesting appearances. If you are curious about the Golden Mean go here, http://en.wikipedia.org/wiki/Golden_ratio this is a truly remarkable explanation.

8. ## digesting your information

I followed the explanation for equation #1. I understand it perfectly. THANK YOU!

Equation #2.
I need to solve for w (not l). But I should be able to get there by the example you have given me. I will give it a shot.
Thanks for the website link -- the question is further in depth and I need to understand the "golden rectangle". So i'm sure it will come in handy!

You have been a tremendous help!