Sum of the Roots
The polynomial y = 3 x 6 − 4 x 4 − 3 x 2 + 2 intercepts the coordinate axis in five places:
(a,0) (b,0) (c,0) (d,0) and (0,e)
What is the value of a + b + c + d + e ?
The answer is 2.
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6 solutions
Assume that y=f(x) so f(-x) = 3(-x)^6 -4(-x)^4 -3(-x)^2 +2 = f(x) when f(-x) = f(x), it means that the function is symetrical by y-axis or we call this as an even function
so... the sum of the solution in x-axis will be zero (a+b+c+d =0)
Next, we must find the value of e e is the value of function when x=0 f(0) = 3(0)^6 -4(0)^4 -3(0)^2 +2 = 2 so we know that e=2 Then a+b+c+d+e = 2
Sorry for teh bad english Thank you for the attention
The equation is symmetrical to y-axis. Therefore, it will intercept x-axis 2n times. n points will be on the positive side and others will be on the negative side. Here n=2 and a = (-d) and b = (-c). And when x=0, y=2 means e = 2. So, a+b+c+d+e=2.
The question is simplified into: What is the sum of the sum of the roots and the y-intercept.
Using Vieta's Theorem, the sum of the roots a + b + c + d = 0 and e = 2 when x = 0. Hence a + b + c + d + e = 0 + 2 = 2.
I have to disagree, because this is a polynomial of the 6th degree, so it has 6 roots, so how can you state that the sum of just 4 of them is zero? The 2 other roots are either complex or some of the a, b, c, d is a multiple root and in either case one can't use Vieta's Theorem to get a+b+c+d=0 (although it is true in this case). :)
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If u can give me a link to vieta's theorem maybe i can tell u but: Since we can turn the equation to 3 x 3 − 4 x 2 − 2 x + 2 so for each of the roots of this equation( s 1 , s 2 , s 3 ),the roots to the main equations are + s 1 , + s 2 , + s 3 , − s 1 , − s 2 , − s 3 The found roots are unacceptable when they are not real but in every case when one of the first equation's roots are positive they give two answers except for when that root is 0. so either way the sum of the roots is 0
A, b, c e d são raízes da equação.
E pela equação de Girard, as soma das raízes é dada pela equação:
-b/a ,como b=0, só nos resta o ponto "e", que quando x=0, sobra e=2
Sum of the roots of the polynomial is 0 as can be easily seen from the polynomial and if x=0 then y=2. Thus a+b+c+d+e=2
Notice that y is an even function, meaning y ( − x ) = y ( x ) . Since there are four roots, they must be arranged symmetrically about the y-axis for y to be an even function. This implies that a + b + c + d = 0 . As for the y-intercept, we simply plug in x=0 and we see that y ( 0 ) = 2 (really, you could just look at the constant term). Therefore, a + b + c + d + e = 0 + 2 = 2 .