Math Help - Difficult simplification problem

1. 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,

2. 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?

3. 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.

4. 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?

5. 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.

6. 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 ?

7. Re: Difficult simplification problem

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!

8. 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.