Rationalizing Square Roots: 27+9t-3t-t^2/9-3t^2-3t+t

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9-t/3-squared t x 3 + square t/3+square t

i get on top 27+9square t -3t-t square t/9-3square t- 3 square t + t

i forgot what happens when u rationa,ize sqaure roots like this
 
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i need help man, anyone know how to add square roots such as 3 square root 5 + 3 square root 7,
 
this doesn't make sense because whenever i plug in the 9 i get 3 and the answer is 6, it doesn't make sense.
 
What does the first post say, it's not clear?
 
Actually NONE of these make sense!
afcwestwarrior said:
9-t/3-squared t x 3 + square t/3+square t

i get on top 27+9square t -3t-t square t/9-3square t- 3 square t + t

i forgot what happens when u rationa,ize sqaure roots like this
If you aren't going to use Latex at least use parentheses to clarify- and "t^2" is clearer than "square t".
I think you mean (9- 3t)/(3-3t^2)+ t^2/(t^2+3). Can you factor 3-3t^2? Get the lowest common denominator and add.

afcwestwarrior said:
i need help man, anyone know how to add square roots such as 3 square root 5 + 3 square root 7,
3\sqrt{5}+ 3\sqrt{7}? About the only thing you can do is factor out the three: 3(\sqrt{5}+ \sqrt{7}). There is no way to "add square roots" of different numbers.

afcwestwarrior said:
this doesn't make sense because whenever i plug in the 9 i get 3 and the answer is 6, it doesn't make sense.
Plug 9 into what? What are you talking about?
 
i figured this one out already, i didn't mean 3 squared i meant the square root of 3
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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