how do you solve this quartic?
Scroll to see replies
Original post by RMNDK
Ok this is a really crazy and possibly stupid way to do this...
But, do prime factor decomposition of 1680.
If you notice, your factors are (x+a), where a is an odd number. This means value of the brackets must be even. Using your factors from factor decomposition, form even factors. You can then cross-reference this to your brackets and figure out what value of x it is.
(Sounds horrible)
But, do prime factor decomposition of 1680.
If you notice, your factors are (x+a), where a is an odd number. This means value of the brackets must be even. Using your factors from factor decomposition, form even factors. You can then cross-reference this to your brackets and figure out what value of x it is.
(Sounds horrible)
Gets an answer extremely quickly for this question even if it feels somewhat inelegant
Original post by morgan8002
I did it by saying b = x - 1, then using quadratic formula on the quadratic in b^2. Gives you answers but not looking nice to simplify so far.
why b = x - 1
what is the motivation?
Original post by TeeEm
why b = x - 1
what is the motivation?
what is the motivation?
That gives you two brackets which are difference of two squares. This simplifies the quartic to a quadratic.
edit: I found b = x - 1 by looking at the values added to x in each bracket. They have a common difference and average to get -1.
(edited 8 years ago)
Original post by TeeEm
I am sure they are because I checked it, but this is a non calculator question .... your number look a bit scary.
Hey, I did all that without a calculator
Original post by BuryMathsTutor
Quick trial and improvement.
x=8? 1 x 5 x 13 x 9 ? No 13 is not a factor of 1680.
x=9? 2 x 6 x 14 x 10 = .
x=8? 1 x 5 x 13 x 9 ? No 13 is not a factor of 1680.
x=9? 2 x 6 x 14 x 10 = .
this was suggested earlier and it will work reasonably quick if the question indicated/suggested integer solutions.
the question is non calculator and says find the two real roots of this equation.
Original post by TeeEm
this was suggested earlier and it will work reasonably quick if the question indicated/suggested integer solutions.
the question is non calculator and says find the two real roots of this equation.
the question is non calculator and says find the two real roots of this equation.
If they weren't integer solutions, it wouldn't be non-calculator.
Just sayin'
Original post by TeeEm
this was suggested earlier and it will work reasonably quick if the question indicated/suggested integer solutions.
the question is non calculator and says find the two real roots of this equation.
the question is non calculator and says find the two real roots of this equation.
too.
I see u = x -1 has already been suggested. I was looking why there's a difference of 4 and thought maybe things will cancel out nicely. (-1 is the mean of -7, -3, 1, 5).
Substituting gives (u-2)(u+2)(u+6)(u-6) after being rearranged, giving u^4 - 40u^2 + 144 = 1680, then you can use the quadratic formula and square root that answer and undo the substitution.
Substituting gives (u-2)(u+2)(u+6)(u-6) after being rearranged, giving u^4 - 40u^2 + 144 = 1680, then you can use the quadratic formula and square root that answer and undo the substitution.
(edited 8 years ago)
Original post by TeeEm
why b = x - 1
what is the motivation?
what is the motivation?
I followed through the working. Most of it's factorisation of numbers but the final result comes out nicely.
some very good techniques and ideas
this question is worthy to go into one of my SPECIALS
I "found" a different way which works because I made the question to work that way ...
I would multiply the brackets in pairs but in a very specific way because you are meant to notice something
the b = x-1 only works here I think because in a desperate attempt to get a reduced quartic (without x^3) you get the unexpected bonus of a quadratic...
Original post by rcmehta
Substitution with u = x - 1 worked out quite well.
u^4 - 40u^2 - 1536 = 0
(u^2-20)^2 = 1936
u^2 = 20 +/- 44
x is real so u^2 = 64
u = +/-8
x = 9, -7
u^4 - 40u^2 - 1536 = 0
(u^2-20)^2 = 1936
u^2 = 20 +/- 44
x is real so u^2 = 64
u = +/-8
x = 9, -7
Seems like I wasted a lot of time with the longer method
Cant u just factor the rhs? In two ways with pos and neg
Posted from TSR Mobile
Posted from TSR Mobile
Original post by rcmehta
Substitution with u = x - 1 worked out quite well.
u^4 - 40u^2 - 1536 = 0
(u^2-20)^2 = 1936
u^2 = 20 +/- 44
x is real so u^2 = 64
u = +/-8
x = 9, -7
u^4 - 40u^2 - 1536 = 0
(u^2-20)^2 = 1936
u^2 = 20 +/- 44
x is real so u^2 = 64
u = +/-8
x = 9, -7
Ahhhh
Note to self: don't trust Google solutions
Posted from TSR Mobile
Original post by TeeEm
THE END
Sorry to continue..
I tried to PM this http://burymathstutor.co.uk/TM.html to you.
Just wanted a few fresh equations to try. Just a bit of fun.
Original post by BuryMathsTutor
Sorry to continue..
I tried to PM this http://burymathstutor.co.uk/TM.html to you.
Just wanted a few fresh equations to try. Just a bit of fun.
I tried to PM this http://burymathstutor.co.uk/TM.html to you.
Just wanted a few fresh equations to try. Just a bit of fun.
your program works (my IT skills are rubbish)
this has the same structure like mine
(x2+18x + ....)(x2+18x+ ...)= ...
Original post by TeeEm
the b = x-1 only works here I think because in a desperate attempt to get a reduced quartic (without x^3) you get the unexpected bonus of a quadratic...
It turns out that when you make the translation it's easy to directly write it in terms of y^2 using difference of 2 squares, and that's maybe a little serendipitous. But the motivation is something I think you can "see" from the original equation.
Original post by DFranklin
Not sure this is really fair. My observation (without really needing to do any calculation) was that a translation y=x+a to the "midpoint" would leave the LHS as an even function of y, at which point we know the quartic is actually a quadratic in y^2.
It turns out that when you make the translation it's easy to directly write it in terms of y^2 using difference of 2 squares, and that's maybe a little serendipitous. But the motivation is something I think you can "see" from the original equation.
It turns out that when you make the translation it's easy to directly write it in terms of y^2 using difference of 2 squares, and that's maybe a little serendipitous. But the motivation is something I think you can "see" from the original equation.
perhaps you have a point.
I am just seeing the question from a different point of view (I made it to work with a particular substitution) and put it out to test what people see. Sometimes when you see how the question works you get s skewed view of things.
Quick Reply
Related discussions
- A-level Maths Question
- PSA: Do NOT buy the ti-84 for A-Levels
- trig identities topic assessment (integral maths) Q5
- inequality involving modulus
- Further Maths AS-level complex numbers question
- Help with integration question
- Maths help: Second derivative differentiation
- Extra MAT 2023
- polar coordinate question help
- Trigonometric Identities
- Pearson Edexcel Further Maths Core Pure 15/5/2023 Topics
- Core pure maths Series question
- Self-teaching A-Level Further Mathematics in One Month Blog
- AQA Level 2 Further Maths
- How to revise maths w pmt
- How do I show aptitude at problem solving if I have only got bronzes in olympiads?
- How should I go about preparing for the STEP Maths exam?
- Computer Science NEA - Advice Needed!
- Oxbridge mathematics supercurriculars, 2025 entry
- LSE Pre-Arrival Proccess
Latest
Trending
Last reply 2 days ago
Did Cambridge maths students find maths and further maths a level very easy?Last reply 2 weeks ago
Edexcel A Level Mathematics Paper 2 unofficial mark scheme correct me if wrongMaths
71
Trending
Last reply 2 days ago
Did Cambridge maths students find maths and further maths a level very easy?Last reply 2 weeks ago
Edexcel A Level Mathematics Paper 2 unofficial mark scheme correct me if wrongMaths
71
