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By comparing co-effs, we can find that sum of the roots is -1*coefficient of x^3 = -4 _________________
"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."
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
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? _________________
"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."
What is the sum of the reciprocals of the roots? _________________
"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."
=(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 _________________
"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."
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
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