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
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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?
Usually I am able to but this one just baffles me
Here's a starter for you: (this one is for the variable t)
Now can you solve for the variable t?
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) ?
Yes. That is correct.
Now can you solve for yourself for the variable r?
would r=3t^2 +or- the square root of (-3t+2t)^2-4(-3t^2)(5r^2+2) over 2(-3t^2) ?
No. It would help if you showed some working as well so we can see where you went wrong.
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)
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
ok lets try with LaTeX
You factorised wrong. Can you see where you went wrong?
not really :/
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.
oh...so it should instead be so then the ? so in that case would the and be combined into ? or would they just be separated to create
Last edited by MizzCyanide; October 6th 2010 at 09:17 PM.
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