rearrange equation

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May 18th 2008, 06:45 AM
Turple

rearrange equation

How would you rearrange:

v=1/3x²h to make x the subject?

Thanks
May 18th 2008, 06:46 AM
Moo

Hello,

Is it $v=\frac 13 \cdot x^2h$ or $v=\frac{1}{3x^2h}$ ?
May 18th 2008, 06:51 AM
Turple

It's a third :)

Also found another 2:

1. It's just the quadratic formula without the +- thing to make c the subject.
2. S=2пr²+2пrh to make h the subject

:)

Thanks
May 18th 2008, 10:08 AM
Simplicity

$v=\frac 13 \cdot x^2h$

Multiply both sides by $3$ :

$3v = x^2 h$

Divide both sides by $h$ :

$\frac{3v}{h} = x^2$

Square root both side:

EDIT: $\pm \sqrt{\frac{3v}{h}} = x$
May 18th 2008, 10:09 AM
Moo

Quote:

Originally Posted by Air

[...]

Square root both side:

${\color{red}\pm} \sqrt{\frac{3v}{h}} = x$

Be careful :p
May 18th 2008, 10:20 AM
Simplicity

Quote:

Originally Posted by Turple

It's a third :)

Also found another 2:

1. It's just the quadratic formula without the +- thing to make c the subject.
2. S=2пr²+2пrh to make h the subject

:)

Thanks

1.)

$x = \frac{-b \sqrt{b^2 - 4ac}}{2a}$

$2ax = -b \sqrt{b^2 - 4ac}$

$-\frac{2ax}{b} = \sqrt{b^2 - 4ac}$

$\left(-\frac{2ax}{b}\right)^2 = b^2 - 4ac$

$\left(-\frac{2ax}{b}\right)^2 - b^2 = -4ac$

$\frac{(-\frac{2ax}{b})^2 - b^2}{-4a} = c$

$c = \frac{b^2 - (\frac{2ax}{b})^2}{4a}$

2.)

$S=2 \pi$ $r^2+2 \pi rh$

$S - 2 \pi r^2 = 2 \pi rh$

$\frac{S - 2 \pi r^2}{2 \pi r} = h$
May 18th 2008, 10:23 AM
Simplicity

Quote:

Originally Posted by Moo

Be careful :p

Edited. (Blush)
May 18th 2008, 01:56 PM
Bradley

collating terms

"How would you rearrange:
v = 1/3 x² h to make x the subject?"

Having seen some of the replies, I wonder if I am missing something here. Do you mean how to I arrange to solve for x?

Assuming you meant (1/3)x and not 1/(3x), first multiply every term by 3 to eliminate fractions:
3v = 3(1/3)x² h
So:
3v = x2 h
Or since you want to isolate x:
x² h = 3v
Divide by coefficient of x:
(x² h)/h = 3v/h
Giving:
x² = 3v/h
Transform by taking square root of both sides:
x = +sqrt(3v/h) and –sqrt(3v/h)

The 1/(3x) would have been more interesting but just as easy.

I see this is the second answer to your question.

I replied because I thought you might be interested in a quick way to isolate any variable in a polynomial if it is at all possible, called the right-to-left method, or sometimes the divide-and-conquer method.

Considering the simplicity of your question, I assume you are probably at the start of your algebra life.

Some polynomials are not amenable to simple algebraic solutions (but there are other methods). The problems you will encounter in beginning algebra or a second course in standard algebra will probably not include such polynomials.

There, I posted something.