Either Complex or Real roots
If f ( x ) = x 5 + x 2 + 1 has roots x 1 , x 2 , x 3 , x 4 , x 5 and g ( x ) = x 2 − 2 , then evaluate
g ( x 1 ) g ( x 2 ) g ( x 3 ) g ( x 4 ) g ( x 5 ) − 3 0 g ( x 1 x 2 x 3 x 4 x 5 ) .
The answer is 7.
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2 solutions
Thanks for the solution.
One doubt which remains is that the given f(x) has only 3 real roots as graph cuts any axis parallel to x-axis at at most 3 points, so how is it possible to have 5 roots?
Should we talk about roots' nature , number of roots- in such type of questions ?
Initially I thought this problem is incorrect because we cannot find the remaining 2 Complex roots of f(x) and substitute it to get first part of the problem.
So please clarify it.
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Roots can be complex. The other two roots are complex and do not shown on graph,
Relevant wiki: Vieta's Formula Problem Solving - Basic
y = g ( x 1 ) g ( x 2 ) g ( x 3 ) g ( x 4 ) g ( x 5 ) − 3 0 g ( x 1 x 2 x 3 x 4 x 5 ) = n = 1 ∏ 5 ( x n 2 − 2 ) − 3 0 g ( − 1 ) = n = 1 ∏ 5 ( x n − 2 ) ( x n + 2 ) − 3 0 ( ( − 1 ) 2 − 2 ) = ( − f ( 2 ) ) ( − f ( − 2 ) ) + 3 0 = f ( 2 ) f ( − 2 ) + 3 0 = ( 4 2 + 2 + 1 ) ( − 4 2 + 2 + 1 ) + 3 0 = ( 3 + 4 2 ) ( 3 − 4 2 ) + 3 0 = 9 − 3 2 + 3 0 = 7 By Vieta’s formula, x 1 x 2 x 3 x 4 x 5 = − 1 Note that f ( x ) = n = 1 ∏ 5 ( x − x n )