Reducing Sqrt(4t^2+4+1/t^2): Calculus Problem Help

AI Thread Summary
To reduce sqrt(4t^2 + 4 + 1/t^2) to (1 + 2t^2)/t, first factor out 1/t^2, leading to (4t^4 + 4t^2 + 1)/t^2. Recognize that 4t^4 + 4t^2 + 1 can be expressed as (2t^2 + 1)^2. This allows for simplification under a single fraction. Ultimately, taking the square root yields the desired result of (1 + 2t^2)/t.
snoggerT
Messages
183
Reaction score
0
reducing sqrt(4t^2+4+1/t^2) to (1+2t^2)/t




The Attempt at a Solution


- This is actually just a portion of a calculus problem, but I can't figure out how the book did the algebra here. I get (2t^2+2t+1)/t and don't know how that reduces to (1+2t^2)/t. Please help.
 
Physics news on Phys.org
Factor the 1/t^2 out as 1/t. Then 4*t^4+4*t^2+1=(2*t^2+1)^2.
 
Put the whole thing under a single fraction : (4t^4 + 4t^2 + 1) / t^2
Now above you have a complete square so you can simplify this and when you do the square root you get the exact solution
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top