x^2 + 4x + 4/x + 1/x^2 + 5= 0 . What is the sum of the roots : PS Archive
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# x^2 + 4x + 4/x + 1/x^2 + 5= 0 . What is the sum of the roots

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x^2 + 4x + 4/x + 1/x^2 + 5= 0 . What is the sum of the roots [#permalink]

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26 Apr 2006, 07:33
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x^2 + 4x + 4/x + 1/x^2 + 5= 0 . What is the sum of the roots of this equation?!

Last edited by laxieqv on 26 Apr 2006, 08:37, edited 1 time in total.
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26 Apr 2006, 08:01
(x-a)(x-b)(x-c)(x-d) = x^4+4x^3+5x^2+4x+1

By comparing co-effs, we can find that sum of the roots is -1*coefficient of x^3 = -4
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26 Apr 2006, 08:32
giddi77 wrote:
By comparing co-effs, we can find that sum of the roots is -1*coefficient of x^3 = -4

Very good

This wasn't easy since two of the roots aren't real.
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26 Apr 2006, 08:36
giddi77 wrote:
(x-a)(x-b)(x-c)(x-d) = x^4+4x^3+5x^2+4x+1

By comparing co-effs, we can find that sum of the roots is -1*coefficient of x^3 = -4

As usual, you nail it, buddy!!! ^_^

Now, my question is: what are the roots of the equation?
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26 Apr 2006, 08:52
i did it bit differently but a bit longer than whats explained above but can be done easily within 2 min ...(well depends )

i took it in this format

x^2 + (1/x^2) +2 +4x + (4/x) +3

= (x + (1/x)) ^2 + 4(x +(1/x) + 3

take x +(1/x) as A

so the above equation changes to A ^2 +4A +3

= (A+1)(A+3)

so A=-1 or A =-3 (infact from here only u can say that the sum is -4 because u can take one root as X and other as 1/x and we want to find the sum of the roots so one value of x +1/x gives the root of one equation and the other gives the value for the other so its sufficient to solve till here but just incase to solve completely we can verify from the following )

but to proceed futher u can sayfor A=-1

x +(1/x) = -1

so x ^2 +x +1 = 0 say the roots are x1 and x2
so x1 + x2 = -b/a = -1

for A =-3

x +(1/x) = -3
so x^2 + 3x +1 =0 say the roots are x3 and x4

so x3 + x4 = -b/a = -3

xo x1 +x2 +x3 +x4 = -3 + -1 = -4
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26 Apr 2006, 08:55
laxieqv wrote:
Now, my question is: what are the roots of the equation?

-(-1)^(1/3),

(-1)^(2/3),

(1/2)*(-3 - sqrt(5)) and

(1/2)*(-3 + sqrt(5)) .

But I'm happy that's not GMAT stuff
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26 Apr 2006, 08:58
laxieqv wrote:
giddi77 wrote:
(x-a)(x-b)(x-c)(x-d) = x^4+4x^3+5x^2+4x+1

By comparing co-effs, we can find that sum of the roots is -1*coefficient of x^3 = -4

As usual, you nail it, buddy!!! ^_^

Now, my question is: what are the roots of the equation?

since the two equations are x^2 +x +1 and X ^2 +3x +1

we can use x = (-b +- sqrt( b^2 -4ac)) /2a

for the firstequation both the roots are imaginary
and for the second equation the roots are (-3 + sqrt(5)) /2 and (-3 - sqrt(5)) /2
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26 Apr 2006, 09:06
ipc302 wrote:
i did it bit differently but a bit longer than whats explained above but can be done easily within 2 min ...(well depends )

i took it in this format

x^2 + (1/x^2) +2 +4x + (4/x) +3

= (x + (1/x)) ^2 + 4(x +(1/x) + 3

take x +(1/x) as A

so the above equation changes to A ^2 +4A +3

= (A+1)(A+3)

so A=-1 or A =-3 (infact from here only u can say that the sum is -4 because u can take one root as X and other as 1/x and we want to find the sum of the roots so one value of x +1/x gives the root of one equation and the other gives the value for the other so its sufficient to solve till here but just incase to solve completely we can verify from the following )

but to proceed futher u can sayfor A=-1

x +(1/x) = -1

so x ^2 +x +1 = 0 say the roots are x1 and x2
so x1 + x2 = -b/a = -1

for A =-3

x +(1/x) = -3
so x^2 + 3x +1 =0 say the roots are x3 and x4

so x3 + x4 = -b/a = -3

xo x1 +x2 +x3 +x4 = -3 + -1 = -4

Very good solution. Shows how math savvy you are!

From my analysis it appears that For any polynomial with a1*x^n+a2*x^(n-1)+a3*x^(n-2)+......+an,
Sum of the roots = -a2/a1

Laxie and ccax.. do you agree?
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26 Apr 2006, 09:07
Now one more Q for all of you:

What is the sum of the reciprocals of the roots?
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26 Apr 2006, 09:19
giddi77 wrote:
Now one more Q for all of you:

What is the sum of the reciprocals of the roots?

say the equation is ax^2 +bx +c = 0

and roots are u and v then

sum of recipocal of roots means

1/u + 1/v

=(u +v ) /uv since sum of roots is always -b/a and product is c/a

= (-b/a) /(c/a)

= -b/c
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26 Apr 2006, 09:34
ipc302 wrote:
giddi77 wrote:
Now one more Q for all of you:

What is the sum of the reciprocals of the roots?

say the equation is ax^2 +bx +c = 0

and roots are u and v then

sum of recipocal of roots means

1/u + 1/v

=(u +v ) /uv since sum of roots is always -b/a and product is c/a

= (-b/a) /(c/a)

= -b/c

You are right. Also after multiplying all the co-efficeints we can see (assuming a,b,c,d are the roots of the polynomial)

I. acbd = 1 (co-efficient of the constant term)
II. abc+bcd+acd+abd = -1*co-efficient of x = -4

Divide eqn II by I, we get
1/a+1/b+1/c+1/d = -4
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26 Apr 2006, 10:03
correct especially in the above equation the roots are x and 1/x for both the equation (i.e they are recipocal of each other ) so in that case the sum of their recipocal will be remain the same so -4
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27 Apr 2006, 03:25
a nice discussion

ipc302's solution is what I expected and it's the typical way to find roots of this kind of symmetric equation!! I deliberately shifted "5" to the back in order to misguide you from the symmetric characteristics of this kind of equation
27 Apr 2006, 03:25
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