Math Help - Can you help me solve this equation?

1. Can you help me solve this equation?

1. Can the formula be solved for either r or t? Explain. 2. if the formula can be solved for r, what does r=? 2-3rt+t=3t^2-t+r^2t^2-5r^2

2. Yes, it can be solved for both r and t.

Do you know how to rearrange the equation into the form of ax^2 + bx + c = 0 and put it onto the quadratic formula?

3. Usually I am able to but this one just baffles me

4. Here's a starter for you: (this one is for the variable t)

$2-3rt+t=3t^2-t+r^2t^2-5r^2$

$2-3rt+t-3t^2+t-r^2t^2+5r^2=0$

$-3t^2-r^2t^2-3rt+2t+5r^2+2=0$

$t^2(-3-r^2)+t(-3r+2)+5r^2+2=0$

Now can you solve for the variable t?

5. is the answer t=3r+2 +or- the square root of (-3r+2)^2-4(-3-r^2)(5r^2+2) over 2(-3-r^2) ?

6. Yes. That is correct.

Now can you solve for yourself for the variable r?

7. would r=3t^2 +or- the square root of (-3t+2t)^2-4(-3t^2)(5r^2+2) over 2(-3t^2) ?

8. No. It would help if you showed some working as well so we can see where you went wrong.

9. I took -3t^2-r^2t^2-3rt+2t+5r^2+2=0 and converted it to r^2(-3t^2)+r(-3t+2t)+5r^2+2=0 and then turned that into r=3t^2 +or- the square root of (-3t+2t)^2-4(-3t^2)(5r^2+2) over 2(-3t^2)

10. Originally Posted by MizzCyanide
I took -3t^2-r^2t^2-3rt+2t+5r^2+2=0 and converted it to r^2(-3t^2)+r(-3t+2t)+5r^2+2=0
I think you factorised wrong... maybe you should try again? As I can see from the equation, there are two r^2 parts. So where'd the second part go?

Also it may pay to learn LaTeX so it is easier to understand.
LaTeX Tutorial

11. ok lets try with LaTeX

I took $-3t^2-r^2t^2-3rt+2t+5r^2+2=0$
$r^2(-3t^2)+r(-3t+2t)+5r^2+2=0$
$r=3t^2 +-\sqrt(-3t+2t)^2-4(-3t^2)(5r^2+2)/2(-3t^2)$

12. You factorised wrong. Can you see where you went wrong?

13. not really :/

14. $r^2(-3t^2)+r(-3t+2t)+5r^2+2=0$

Now can you see that 5r^2 has an r squared variable in it? It should be factorised along with the other r squared variable.

15. oh...so it should instead be $r^2(5+2)$ so then the $(5+2)=7$? so in that case would the $(-3t^2)$ and $(5+2)=7$ be combined into $(-3t^2+7)$? or would they just be separated to create $r=3t^2+-\sqrt(-3t+2t)^2-4(-3t^2)(7)/2(-3t^2)$

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