Polynominals (1 Viewer)

Doctor Jolly

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Help?

I have no idea how to do one of these questions!

The product of two of the roots of
x^4 + 2x^3 - 18x - 5 = 0
is -5. Find the product of the other two roots.

Thanks in advance!
 

conics2008

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ok since its to the power of 4 there has to be atleast 4 roots

do this let alpha x beta = -5 ...

let me do it on paper then i will post it up =)
 

foram

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Doctor Jolly said:
Help?

I have no idea how to do one of these questions!

The product of two of the roots of
x^4 + 2x^3 - 18x - 5 = 0
is -5. Find the product of the other two roots.

Thanks in advance!
product of all 4 roots = e/a = -5

product of the first 2 roots is -5, therefroe the product of the other two roots is 1. I think.
 

conics2008

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lol.. i forgot about you lol...

let a b c d be the roots

let a x b = -5

the product of 4 roots = abcd= e/a

since you said a x b = -5

therefore -5cd=-5

divide by -5 => cd=1

therefore the other product are 1

=)

sorry i completly forgot about you
 

foram

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I thought that if I finnish 4U math this year, then thats less work for me next year. And i'll have more revision time. :D
 

conics2008

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how on earth do you do it by your self.

last year i went ahead of the class for only 3unit and 2unit never thought about starting 4unit ...


dont you think its too much =)
 

foram

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conics2008 said:
dont you think its too much =)
It's not that much work, besides, since i've finnished prelim phys and chem last week, i'll have lots of spare time. :D
 

conics2008

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wtfff.. you finished ( with your class ) or on your own the whole subject...

wow you guys are really fast..

our one everybody is at the same level. but i prefer to go ahead only in maths. maths is a cool subject.. sometimes u gotta change the question to suit the answer =) lol.. i did this once

I onced changed an angel to suit the answer lol....
 

foram

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conics2008 said:
wtfff.. you finished ( with your class ) or on your own the whole subject...

wow you guys are really fast..

our one everybody is at the same level. but i prefer to go ahead only in maths. maths is a cool subject.. sometimes u gotta change the question to suit the answer =) lol.. i did this once

I onced changed an angel to suit the answer lol....
I learnt prelim phys and chem on my own. It's not very much work considering how it's supposed to take 3 terms.
 

someauzzie

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wow i feel really dumb now.... im jus goin with the flow in class.....
 

tommykins

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Doctor Jolly said:
Help?

I have no idea how to do one of these questions!

The product of two of the roots of
x^4 + 2x^3 - 18x - 5 = 0
is -5. Find the product of the other two roots.

Thanks in advance!
Let roots be a,b,c,d

Product of roots = abcd = -5cd = -5

cd = 1
 

Doctor Jolly

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oh my.
Thanks for the overwhelming response guys!
I finally get it, I can't believe I didn't before ;D
 

Doctor Jolly

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oh great, i don't get something else.. new question:

find the value of k if the quadratic equation:
x^2 - (k+2)x + k + 1 = 0 has

i. consecutive roots. answer: k = - 1 and +1

ii. equal roots. answer: k = 0

iii. one root double the other answer: k = -0.5 and 1


wargh.. i hate this bit of polynominals ><" :mad:
 

foram

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Doctor Jolly said:
oh great, i don't get something else.. new question:

find the value of k if the quadratic equation:
x^2 - (k+2)x + k + 1 = 0 has

i. consecutive roots. answer: k = - 1 and +1

ii. equal roots. answer: k = 0

iii. one root double the other answer: k = -0.5 and 1


wargh.. i hate this bit of polynominals ><" :mad:
EDIT: for i.

Let the roots be A, A+1

-b/a = k+2 = 2A + 1
c/a = k+1 = A(A+1)

k+2-k-1 = 2A+1-A(A+1)
2A+1-A^2 -A = 1
A-A^2=0
A=1 , 0

P(0) = k +1 =0
k=-1

I don't know how they got k= 1
 
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Doctor Jolly

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hm, i don't know how to do it. The answers are the ones at the back of the book btw.
 

tommykins

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Doctor Jolly said:
oh great, i don't get something else.. new question:

find the value of k if the quadratic equation:
x^2 - (k+2)x + k + 1 = 0 has

i. consecutive roots. answer: k = - 1 and +1

ii. equal roots. answer: k = 0

iii. one root double the other answer: k = -0.5 and 1


wargh.. i hate this bit of polynominals ><" :mad:
.
i. Let roots be a and a+1
a+1 + a = k+2
2a + 1 = k+2
k = 2a-1 equation (1)

a²+a = k+1
k = a²+a-1 equation (2)

2a - 1 = a²+a-1
a²-a = 0
a = 0 or 1.

Sub into 91)
k = 2(0) - 1 = -1
k = 2(1) - 1 = 1


ii. Discriminant (k+2)² - 4(k+1) = 0 => k²+4k+4 - 4k -4 = 0 => k² = 0 .:. k = 0

iii. Try lettnig roots be a and 2a.
3a = k+2
k = 3a-2 equation (1)

2a² = k+1
k = 2a²-1 equation (2)

3a-2 = 2a² -1
2a² - 3a + 1 = 0
a = 1, 0.5

Sub these into the k = 3a -2
k = 3(1) - 2 = 1
k = 3(0.5) - 2 = 1.5 - 2 = -0.5
iii.
 
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foram

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oh. I see now. I was supposed to sub the value I got for the root into the equation I got using the relationship between the roots and the coefficients :D... you're smart tommykins. :D
 

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