CSSA 2008 - Polynomials 4d (1 Viewer)

[Tsylana]

I cant seem to get this question down and i'm not really sure why ><" tried it for like 30 minutes and I couldn't get it would someone be able to post up a solution?

The polynomial equation x^5 - ax^2 + b = 0 has a multiple root.

Show that 108a^5 = 3125b^3

 

[azureus88]

[maths]f(x)=x^5-ax^2+b\\f'(x)=5x^4-2ax\\let\,root\,be\,x=\alpha \\f'(\alpha)=5\alpha^4-2a\alpha=0\\\alpha^3=\frac{2a}{5}\,since\,\alpha\neq 0\\f(\alpha)=\alpha^5-a\alpha^2+b=0\\\alpha^2(\alpha^3-a)+b=0\\\alpha^2(\frac{2a}{5}-a)+b=0\\\alpha^2=\frac{5b}{3a}\\\\\alpha(\frac{5b}{3a})=\frac{2a}{5}\\\alpha=\frac{6a^2}{25b}\\\\f'(\alpha)=5\alpha^4-2a\alpha=0\\5(\frac{6^4a^8}{25^4b^4})-2a(\frac{6a^2}{25b})=0\\Simplifying, \\108a^4=3125b^3[/maths]

 

[jet]

This was my trial

 

[kwabon]

[maths]f(x)=x^5-ax^2+b\\f'(x)=5x^4-2ax\\let\,root\,be\,x=\alpha \\f'(\alpha)=5\alpha^4-2a\alpha=0\\\alpha^3=\frac{2a}{5}\,since\,\alpha\neq 0\\f(\alpha)=\alpha^5-a\alpha^2+b=0\\\alpha^2(\alpha^3-a)+b=0\\\alpha^2(\frac{2a}{5}-a)+b=0\\\alpha^2=\frac{5b}{3a}\\\\\alpha(\frac{5b}{3a})=\frac{2a}{5}\\\alpha=\frac{6a^2}{25b}\\\\f'(\alpha)=5\alpha^4-2a\alpha=0\\5(\frac{6^4a^8}{25^4b^4})-2a(\frac{6a^2}{25b})=0\\Simplifying, \\108a^4=3125b^3[/maths]

woops, 108a^5.

 

[Tsylana]

haha wow... . i just did it like 5 seconds before i checked up on here after like 45 minutes of trying it but i got it doing it really strangely lol. x__X... haha seems theres lots of quirky methods around... i like the first guys method seems quite genius tho i dont think im gonna be able to remember when to use something like that before my trials x___x

 

[jet]

haha wow... . i just did it like 5 seconds before i checked up on here after like 45 minutes of trying it but i got it doing it really strangely lol. x__X... haha seems theres lots of quirky methods around... i like the first guys method seems quite genius tho i dont think im gonna be able to remember when to use something like that before my trials x___x

What's wrong with my method?????

 

[azureus88]

hm...i actually like jetblacks method, much simpler and less subbing in

 

[Timothy.Siu]

yeah his ones good. I wouldn't have thought of doing that. I did it the long subbing in way.

 

[Tsylana]

fundamentally theres nothing wrong with it i just never did anything like that so i wouldnt be able to remember it... and since i have trials the day after tomorrow i dont think i would be able to do something like that.. =="

 

[kwabon]

stick to jetblack's method, easiest way out and many people do it that way, which includes me.

 

[Aerath]

Yup, I prefer jetblack's. That being said, it really doesn't matter which you use - as long as you get to the answer.

 

[cutemouse]

lol, this was in our school's trial for 2003. Wonder where they got it from?

 

[tommykins]

why are you trying to remember methods..