# Rearranging a fixed point equation

• May 24th 2010, 02:12 PM
NitrousUK
Rearranging a fixed point equation
Hi, sorry if this is a silly/simple question! But i've been banging my head against it for a while. There's an example of a fixed point equation in my course book that gives $\displaystyle \frac{-1}{8}x^2+\frac{5}{8}x+\frac{7}{2}=x$

But then rearranges to give $\displaystyle x^2+3x-28=0$

I can see they're dividing through by the 'a' coefficient for the quadratic, but I don't understand the operation involved in turning the 'b' coefficient from 5/8 divide by 1/8, which gives me 5, into the 3 when the RHS (which is 'x' over 1/8) is transposed.

Any help/hints greatly appreciated! Thanks.
• May 24th 2010, 02:18 PM
pickslides
$\displaystyle \frac{-1}{8}x^2+\frac{5}{8}x+\frac{7}{2}=x$

$\displaystyle \frac{-1}{8}x^2+\frac{5}{8}x+\frac{7}{2}=\frac{8}{8}x$

$\displaystyle \frac{-1}{8}x^2+\frac{-3}{8}x+\frac{7}{2}=0$

$\displaystyle x^2+3x+\frac{-56}{2}=0$

$\displaystyle x^2+3x-28=0$
• May 24th 2010, 02:18 PM
e^(i*pi)
Quote:

Originally Posted by NitrousUK
Hi, sorry if this is a silly/simple question! But i've been banging my head against it for a while. There's an example of a fixed point equation in my course book that gives $\displaystyle \frac{-1}{8}x^2+\frac{5}{8}x+\frac{7}{2}=x$

But then rearranges to give $\displaystyle x^2+3x-28=0$

I can see they're dividing through by the 'a' coefficient for the quadratic, but I don't understand the operation involved in turning the 'b' coefficient from 5/8 divide by 1/8, which gives me 5, into the 3 when the RHS (which is 'x' over 1/8) is transposed.

Any help/hints greatly appreciated! Thanks.

If we multiply through by 8 (the LCD of 2 and 8)

$\displaystyle -x^2 + 5x + 28 = 8x$

The 3x comes from $\displaystyle 8x-5x$ when simplifying

$\displaystyle x^2+3x-28 = 0$
• May 24th 2010, 02:23 PM
Also sprach Zarathustra
Quote:

Originally Posted by NitrousUK
Hi, sorry if this is a silly/simple question! But i've been banging my head against it for a while. There's an example of a fixed point equation in my course book that gives $\displaystyle \frac{-1}{8}x^2+\frac{5}{8}x+\frac{7}{2}=x$

But then rearranges to give $\displaystyle x^2+3x-28=0$

I can see they're dividing through by the 'a' coefficient for the quadratic, but I don't understand the operation involved in turning the 'b' coefficient from 5/8 divide by 1/8, which gives me 5, into the 3 when the RHS (which is 'x' over 1/8) is transposed.

Any help/hints greatly appreciated! Thanks.

$\displaystyle \frac{-1}{8}x^2+\frac{5}{8}x+\frac{7}{2}=x$
multiply by 8
$\displaystyle -x^2+5x+28=8x$
multiply by -1

$\displaystyle x^2-5x-28=-8x$
$\displaystyle x^2-5x-28+8x=0$
$\displaystyle x^2-3x-28=0$
• May 24th 2010, 02:27 PM
NitrousUK
Multiply by the denominator to remove the fractions! Of course..