Thread: Weird Basic Algebra Problem (Help Appreciated =D )

1. Weird Basic Algebra Problem (Help Appreciated =D )

Hello,

I got this annoying problem I can't seem to get around, I know there's a trick somewhere in the book for this...I think there is, anyhow, I need to solve the following equation for x:

(s^2 - 2sx + x^2)(a) = (x^2)(b)

Each time I try to mess around with x and s, I'm unable to break the 2sx term and it's rly getting to me, at my last step I got to this useless mess:

(s^2 - 2sx)/(x^2) = (b/a)(x^2)

Any tips/tricks highly appreciated, spanks!

2. Originally Posted by FIRESSS
Hello,

I got this annoying problem I can't seem to get around, I know there's a trick somewhere in the book for this...I think there is, anyhow, I need to solve the following equation for x:

(s^2 - 2sx + x^2)(a) = (x^2)(b)

Each time I try to mess around with x and s, I'm unable to break the 2sx term and it's rly getting to me, at my last step I got to this useless mess:

(s^2 - 2sx)/(x^2) = (b/a)(x^2)

Any tips/tricks highly appreciated, spanks!
Hello,

if you factorize the first bracket you'll notice that it is a complete square: $\displaystyle s^2-2sx+x^2=(s-x)^2$

After a few steps you'll get the proportion:

$\displaystyle \frac{(s-x)^2}{x^2}=\frac{a}{b} \Longleftrightarrow \left( \frac{s-x}{x} \right)^2=\frac{a}{b}$

EB

3. Hello, FIRESSS!

Solve for $\displaystyle x:\;\;a(s^2-2sx+x^2)\;=\;bx^2$

Why not expand it and solve the resulting quadratic equation?

We have: .$\displaystyle as^2 -2asx + ax^2\;=\;bx^2\quad\Rightarrow\quad(a-b)x^2 - 2asx + as^2\;=\;0$

Quadratic Formula: .$\displaystyle x \;=\;\frac{-(\text{-}2as) \pm \sqrt{(\text{-}2as)^2 - 4(a-b)(as^2)}}{2(a-b)}$

. . which simplifies to: .$\displaystyle x \;=\;\frac{as \pm s\sqrt{ab}}{a-b}$