Could you also please show the process? Thank you!
2006-11-02
15:57:43
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7 answers
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asked by
blue skies
2
in
Science & Mathematics
➔ Mathematics
It's a homework assignment.
2006-11-02
16:02:00 ·
update #1
The question just says: "Solve the following equations". So I assume I'm supposed to be solving for x right? Why would the teacher give an impossible equation. :(
2006-11-02
16:09:10 ·
update #2
maybe its a trick question or mistake :P
2006-11-02
16:14:26 ·
update #3
divide by x^2
6(x^2+(1/x^2))+ (x)+(1/x)+10 = 0
put x+(1/x)= t
then solve for t and then solve for x
you'll find that there are no roots.!
2006-11-02 16:03:29
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answer #1
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answered by !kumar! 2
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First of all it is a wrong question. No products of four factors will give 6x^4 + x^3 + 10x^2 + x + 6 = 0 as it doesn't match with the y-intercept i.e 6.
It is possible to solve this type of equation. Many people think only quadratic or linear equations can be solved. Using a proper method it can solved.
Observe: c = 6. That means from general equation
(x+a)(x+b)(x+c)(x+D)
c=a*b*c*d
a*b*c*d=6
take factors of 6 which multiply four times to give c value as 6.
Factors are: -1,-2,-3,-6,1,2,3,6
Only one combination will work. (Select 4 numbers from the list which give product of 6 and divide to give reminder 0. (Long process)
For example: If we take value 1 i.e (x+1). Now divide the main equation by (x+1) to get a cubic equation. If the reminder is 0, then the factor is correct. Then one of the values will be x = -1
2006-11-02 16:30:41
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answer #2
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answered by prashmanic 4
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because of the fact that a number of those are superior mutually, making each area equivalent 0 makes the full equation equivalent 0 If x=6, then (x-6) equals 0, and multiplying 0 by skill of something equals 0. So x equals 6, 17 or -4 is actual. yet which one? in case you in basic terms artwork the undertaking out with any of those values for x, you will get 0. So all are maximum remarkable.
2016-11-27 00:42:01
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answer #3
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answered by Anonymous
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This problem is simple.
look at the coefficients
they are 6,1,10,1,6
it is symetric
so (x+1/x) comes as a part of a factor
devide by x^2
6x^2+x+10+ 1/x+6/x^2
= 6(x^2+1/x^2)+ (x+1/x) + 10
let x+1/x = t or tx = x^2+1
so x^2+2/x^2 = t^2 -2
so expression/x^2 = 6(t^2-2) + t + 10
= 6t^2 +t - 2
= 6t^2+ 4t-3t-2
= 2t(3t+2) -(3t+2)
= (2t-1)(3t+2)
expression = x^2(2t-1)(3t+2)
= (2tx -x)(3tx + 2x)
= (2x^2-2-x)(3x^2+3 + 2x)
= (2x^2-x-2)(3x^2+2x + 3)
both being binomial can be attempted as below
(2x^2-x-2) = (2x+1)(x-2)
(3x^2+2x+3 cannot be factored
so (2x+1)(x-2)(3x^2+2x+3)
2006-11-02 17:54:21
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answer #4
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answered by Mein Hoon Na 7
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impossible
the first 4 terms must equal -6
if x is positive this isnt possible
if x is negative the second and 4th term would be negative but the first and third term are too large respectively to the next term that would be negative. meaning the 6x^4+x^3+10x^2+x has to be positive
2006-11-02 15:59:56
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answer #5
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answered by RichUnclePennybags 4
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impossible, no matter what value there is for x, the number will be no less than six. (0,6)
2006-11-02 16:03:45
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answer #6
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answered by Anonymous
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the answer to your problem is: no solution
2006-11-02 16:20:12
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answer #7
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answered by dew li low moe 2
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