# Difficult simplification problem

• Sep 25th 2012, 11:07 PM
frustrated
Difficult simplification problem
I am trying to simplify this to express a as a function of t,r,p. It is killing me. Any help would be most appreciated.

a^2=[(at)^2 + r^2]/p

Edit: This problem is trivial, but I am also looking for a solution given the additional constraint,

t=ab,
• Sep 25th 2012, 11:19 PM
MarkFL
Re: Difficult simplification problem
We are given to solve for $a$:

$a^2=\frac{(at)^2+r^2}{p}$

The first step is to multiply through by $p$:

$a^2p=(at)^2+r^2$

Now, subtract through by $(at)^2$:

$a^2p-(at)^2=r^2$

$a^2p-a^2t^2=r^2$

Factor the left side:

$a^2(p-t^2)=r^2$

Can you proceed from here?
• Sep 25th 2012, 11:42 PM
frustrated
Re: Difficult simplification problem
Okay, but (at)^2 does not equal a^2t^2, it equals a^2 +2at + t^2

I am stuck trying to get rid of the 2at term.
• Sep 25th 2012, 11:45 PM
MarkFL
Re: Difficult simplification problem
$(at)^2=a^2t^2$ however $(a+t)^2=a^2+2at+t^2$, but this is not what you posted.

Which do you mean?
• Sep 25th 2012, 11:55 PM
frustrated
Re: Difficult simplification problem
Sorry, I was writing and reading (at)^2 and thinking (a+t)^2. The former is the correct expression. Thank you for our help.
• Sep 26th 2012, 01:48 AM
frustrated
Re: Difficult simplification problem
Okay, what if we introduce a further identity,

t=ba

giving

A^2p - B^2A^4 = r^2

Is this soluble ?
• Sep 26th 2012, 02:02 AM
earboth
Re: Difficult simplification problem
Quote:

Originally Posted by frustrated
Okay, what if we introduce a further identity,

t=ba

giving

A^2p - B^2A^4 = r^2

Is this soluble ?

1. I assume

a) that you mean: $a^2 \cdot p - b^2 \cdot a^4 = r^2$

b) that you want to solve for a.

If so:

2. Re-arrange the given equation to

$b^2 \cdot a^4 - p \cdot a^2 + r^2 = 0$

Replace $a^2 = y ~ \implies ~ |a| = \sqrt{y}$

$b^2 \cdot y^2 - p \cdot y + r^2 = 0$

is a quadratic in y which will give the solutions:

$y= \frac{p \pm \sqrt{p^2-4b^2 r^2}}{2b^2}$

3. That means $|a| = \sqrt{\frac{p \pm \sqrt{p^2-4b^2 r^2}}{2b^2}}$

Btw: If you have a new question start a new thread!
• Sep 26th 2012, 02:31 AM
frustrated
Re: Difficult simplification problem
I am confused, because your solution gives a=zero when b=0, but if we simply let b=0 and solve from the initial equation, we get a simple solution that is only zero when r=0, i.e.

a=SQRT(r^2/p)

Edit: I see that the quadratic formula does not apply when b=0.