Quadratic Equation (Check my Workings ?)

AI Thread Summary
The quadratic equation 3t^2 + 7t - 5 = 0 was solved using the quadratic formula. The calculations led to two potential solutions: t = -4.39 and t = -9.61. However, these results differ from the tutor's answer of t = 0.573 and t = -2.907. The error was identified as a miscalculation in the denominator, which should not equal 4. This highlights the importance of careful arithmetic in solving quadratic equations.
lloydowen
Messages
78
Reaction score
0

Homework Statement



Solve using the formula method.

Homework Equations



3t^2+7t=5

The Attempt at a Solution



3t^2+7t-5=0<br /> t= \frac{-7 +or-\sqrt{7^2-4(3)(-5)}}{2(3)}<br /> <br /> t= \frac{-7+\sqrt{109}}{4}<br /> t= -4.39<br /> OR...
t= \frac{-7-\sqrt{109}}{4}<br /> t= -9.61

The reason I have doubts is because the answer book from the tutor shows, T= 0.573 or -2.907... ? :(
 
Physics news on Phys.org
Ah ****, misuse of the formula tags, :( For some reason it won't let me go back and edit, hope you guys can understnad that :/
 
lloydowen said:
3t^2+7t-5=0<br /> t= \frac{-7 +or-\sqrt{7^2-4(3)(-5)}}{2(3)}<br /> <br /> t= \frac{-7+\sqrt{109}}{4}<br /> t= -4.39<br /> OR...
t= \frac{-7-\sqrt{109}}{4}<br /> t= -9.61

The reason I have doubts is because the answer book from the tutor shows, T= 0.573 or -2.907... ? :(
In the denominator, 2(3) does not equal 4.
 
Oh! Such a silly mistake, I feel like a fool now !
 
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
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top