Algebraic Sum Of Roots
If α , β , and γ are the roots of x 3 − x − 1 = 0 , compute:
1 + α 1 − α + 1 + β 1 − β + 1 + γ 1 − γ
The answer is 1.
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5 solutions
Can be done in a much simpler and elegant way !! a,b,c be the roots of the given polynomial, then we have to find S = \frac{1-a}{1+a} + \frac{1-b}{1+b} +\frac{1-c){1+c} , let p {1} = \frac{1-a}{1+a} ,p {2} =\frac{1-b}{1+b} ,p {3} =\frac{1-c}{1+c} . now p {1}= \frac{1-a}{1+a} then a = \frac{1-p {1}}{1+p {1}} now 'a' is a root of the given polynomial therefore , a^{3} -a-1 = 0 -------equation (1) ,therefore , substituting a = \frac{1-p {1}}{1+p {1}} we get p1^{3}-p1^{2}+7p1 +1 = 0 therefore p1, satisfies the equation p^{3}-p^{2}+7p +1= 0....equation (2) , by similar argument, even p {2} and p {3} satisfy equation(2), hence p {1},p {2},p {3} are roots of the equation(2) , now S = p {1} + p {2} + p {3} = 1
An easy and less complicated way for solving the problem........ thanks for this solution...!!!!!!!!
Nice solution. Exactly the same way i did it.
little longer it is!!!
How did the roots α + β + γ become 0?
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sum of the roots (coeff of x2) is 0
becoz there is no value of coefficient of x2
the formula is = -b/a its mean 0/1 = undefined (0)
The denominator should also contain (alpha + beta + gamma)????
The problem can be solved by figuring out the equation whose roots would be (1-a)/(1+a). By simple substitution we can find out that these would be the roots of the equation if x is replaced by (1-x)/(1+x). After we get the equation the solution of the problem is nothing but the -(coeff. of x^2)/(coeff. of x^3) in the new equation
We're given: x ( x − 1 ) = x + 1 1 .
So, we have: 1 + α 1 − α = α ( α − 1 ) ( 1 − α ) = − α 3 + 2 α 2 − α .
Then, adding the β and γ terms, we have:
S = − ( α 3 + β 3 + γ 3 ) + 2 ( α 2 + β 2 + γ 2 ) − ( α + β + γ )
⟹ S = − ( 3 α β γ ) + 2 ( ( α + β + γ ) 2 − 2 α β − 2 β γ − 2 γ α ) − ( 0 )
⟹ S = − ( 3 ( 1 ) ) + 2 ( 2 ) − ( 0 ) = 1
Nice solution
Cool. I learnt a new method today. Thanks.
how did you work out that a 3 + b 3 + c 3 = 3 a b c
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(I mean alpha beta and gamma, but I dunno how to type that in latex)
reply parth
You can use Newton’s sums to find out α 3 + β 3 + γ 3 as well as α 2 + β 2 + γ 2 Being: S 3 = α 3 + β 3 + γ 3 S 2 = α 2 + β 2 + γ 2 S 1 = α + β + γ S 0 = 3 S − 1 = α 1 + β 1 + γ 1 Also, expanding the polynomial: x 3 + 0 x 2 − x − 1 = 0 , you have: 1 . S 3 + 0 . S 2 − 1 . S 1 − 1 . S 0 = 0 S 3 − S 1 − S 0 = 0 α + β + γ = a − b = 0 S 3 = S 0 S 3 = 3 To find out S 2 : 1 . S 2 + 0 . S 1 − 1 . S 0 − 1 . S − 1 = 0 S 2 − S 0 − S − 1 = 0 S 2 = 3 + S − 1 ( I ) S − 1 = α 1 + β 1 + γ 1 S − 1 = α . β . γ β . γ + α . γ + α . β β . γ + α . γ + α . β = a c = − 1 α . β . γ = a − d = 1 S − 1 = 1 − 1 = − 1 Back to ( I ) : S 2 = 3 − 1 S 2 = 2 From Parth’s solution: S = − ( α 3 + β 3 + γ 3 ) + 2 ( α 2 + β 2 + γ 2 ) − α + β + γ S = − S 3 + 2 S 2 − S 0 S = − 3 + 2 . 2 − 0 S = 1
Can you please elaborate on the second line? I feel lost.
A = 1 + α 1 − α + 1 + β 1 − β + 1 + γ 1 − γ
A = ( 1 + α ) ( 1 + β ) ( 1 + γ ) ( 1 − α ) ( 1 + β ) ( 1 + γ ) + ( 1 − β ) ( 1 + α ) ( 1 + γ ) + ( 1 − γ ) ( 1 + α ) ( 1 + β )
On solving this, we get:
A = 1 + ( α + β + γ ) + ( α β + β γ + γ α ) + ( α β γ ) 3 + ( α + β + γ ) − ( α β + β γ + γ α ) − 3 ( α β γ )
which equals:
A = 1 + 0 + ( − 1 ) + 1 3 + 0 − ( − 1 ) − 3 ( 1 )
A = 1
In the given equation substitute (1-t)/(1+t) for 'x' and obtain a new cubic equation in terms of 't' . Then the required expression is simply the sum of the roots of the new cubic....
apply componendo and dividendo and the eq. simplifies to-
= -(1/a+1/b+1/c)
abc=1 $ ab+bc+ca=-1
divide the 2 eq. and you get (1/a+1/b+1/c)
Let the required expression be S . Therefore S = 1 + α 1 − α + 1 + β 1 − β + 1 + γ 1 − γ
S + 3 = ( 1 + α 1 − α + 1 ) + ( 1 + β 1 − β + 1 ) + ( 1 + γ 1 − γ + 1 )
S + 3 = 1 + α 2 + 1 + β 2 + 1 + γ 2
2 S + 3 = 1 + ( α β + β γ + γ α ) + ( α β γ ) 3 + 2 ( α + β + γ ) + ( α β + β γ + γ α )
2 S + 3 = 1 + ( − 1 ) + 1 3 + 2 ( 0 ) + ( − 1 )
2 S + 3 = 2
S = 1