Help on tis quadratic functions question

AI Thread Summary
The discussion revolves around solving a quadratic function related to the speed of a particle, expressed as v = 10t - t². The goal is to demonstrate that the particle maintains a speed of 5 m/s or greater for exactly 4√5 seconds. The initial attempt involved setting up the inequality 10t - t² > 5, leading to the quadratic equation t² - 10t + 5 < 0. The solution reveals that the relevant time interval is between the two roots, calculated as 4√5 seconds, confirming the problem's requirements. The conversation emphasizes the importance of recognizing the solution without getting lost in decimal approximations.
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[SOLVED] Help on tis quadratic functions question

Homework Statement


The speed v m/s of a particle traveling from A to B at time t s after leaving A, is given by v =10t-t². the particle starts from rest at A and comes to rest at B. Show that the particle has a speed of 5m/s or greater for exactly 4\sqrt{5} s



Homework Equations


NOne


The Attempt at a Solution



10t-t²>5
t²-10t+5 <0
0.52 < t < 9.47
then i don't know how to do already.
Can someone start me off and guide me along?
 
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consider using the quadratic formula
 
Sigh. You have already solved the problem, but because you are so infatuated with decimal approximations, you don't see the answer even when it stares you in the eye.

On your second line, we factorize the quadratic, with the two zeroes:
X=\frac{-(-10)\pm\sqrt{(-10)^{2}-4*1*5}}{2*1}=\frac{10\pm\sqrt{80}}{2}=\frac{10\pm\sqrt{16*5}}{2}=5\pm{2}\sqrt{5}

You have found that the relevant region lies between these 2 zeroes, and we may then calculate the distance between them:
(5+2\sqrt{5})-(5-2\sqrt{5})=4\sqrt{5}
as was to be shown.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
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
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