# Math Help - Rearrange in terms of z

1. ## Re: Rearrange in terms of z

Originally Posted by TomByrne
Hi,
I need to rearrange this function in terms of z, I'm struggling a little bit (asterisks are multiplication):

a*(z-b) = -c*d*(f/(f+z))-(c*g)-h*e*(f/(f+z))-(h*i)

Any help?
\displaystyle \begin{align*}a(z - b) &= \frac{cdf}{f + z} - cg - \frac{efh}{f +z } - hi \\ az - ab &= \frac{cdf}{f + z} - \frac{cg(f + z)}{f + z} - \frac{efh}{f + z} - \frac{hi(f + z)}{f + z} \\ az - ab &= \frac{cdf - cg(f + z) - efh - hi(f + z)}{f + z} \\ (az - ab)(f + z) &= cdf - cg(f + z) - efh - hi(f + z) \\ afz + az^2 - abf - abz &= cdf - cgf - cgz - efh - fhi - hiz \\ az^2 + afz + abz + cgz + hiz - abf - cdf + cgf + efh + fhi &= 0 \\ az^2 + \left(af + ab + cg + hi\right)z - abf - cdf + cgf + efh + fhi &= 0 \end{align*}

This is a quadratic of the form $\displaystyle \alpha z^2 + \beta z + \gamma = 0$, which has solution $\displaystyle z = \frac{-\beta \pm \sqrt{\beta ^2 - 4\alpha\gamma}}{2\alpha}$.

2. ## Re: Rearrange in terms of z

Originally Posted by TomByrne
Thanks, great. What are the square brackets in this situation?
SAME as (); simply makes for improved appearance; easier to tell what the numerator is.

3. ## Re: Rearrange in terms of z

Originally Posted by Prove It
\displaystyle \begin{align*}az^2 + \left(af + ab + cg + hi\right)z - abf - cdf + cgf + efh + fhi &= 0 \end{align*}
ProveIT, I think your "ab" should be negative, and your "cdf" should be positive.
Rest of them are same as mine...

4. ## Re: Rearrange in terms of z

Originally Posted by Wilmer

What you think, Pickslide? You proud of me?
Love it, have a beer for me.

5. ## Re: Rearrange in terms of z

Originally Posted by Wilmer
ProveIT, I think your "ab" should be negative, and your "cdf" should be positive.
Rest of them are same as mine...
Well, I agree that ab should be negative. But cdf was positive on the right, so should be negative on the left.

6. ## Re: Rearrange in terms of z

Originally Posted by Prove It
Well, I agree that ab should be negative. But cdf was positive on the right, so should be negative on the left.
No, cdf was negative on the right; original equation:
a * (z-b) = -c * d * (f / (f+z)) - (c * g) - h * e * (f / (f+z)) - (h * i)

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