3 Answers | Add Yours
let f(x) = ax^5 + bx^3 + c
Given that f(x) has 5 roots:
If c = 0 ==> f(x) = x^3(ax^2 + b)
==> f(x) will have 3 roots oly, then c can not be 0.
If b = 0 ==> f(x) = ax^5 + c ==> x = (c/a)^1/5 is the only zero.
Then b can not be zero.
if a = 0 ==> f(x) = bx^3 +c ==> f(x) will be 3 gegree funcion and wil not have 5 roots. then a can not be 0 .
We conclude that:
if f(x) has 5 roots, then a,b, c, can not be zero.
For a function f(x) which has five roots, the highest power of x should be 5. We are given f(x) = ax^5 + bx^3 + c.
So this has to have the highest power of x as 5, therefore 'a' cannot be zero else the expression will have a highest power of 3.
Also c cannot be zero, because if it is taken to be zero, the expression is reduced to ax^5 + bx^3= x^3(ax^2+b), this can have a maximum of three roots including 0.
If b is zero the expression is ax^5+c. Now equating ax^5+c=0,
x^5 = c/a . Now this also cannot yield 5 values of x.
So neither of a , b or c can be zero if the expression has 5 roots.
f(x) = ax^5+bx^3+c has 5 roots.
To determine which of a, b and c could be zero.
If a = 0, then f(x) = bx^3+c becomes a 3rd degree polnomial which could have 3 roots. So it cannot have 5 roots.
So a cannot be zero.
If c = 0, then f(x) = ax^5+bx^3 = 0 becomes x^3(ax^2+b) = so x =0 or 2 roots from ax^2+b = 0 whic contradicts f(x) has 5 roots.
If b= 0, then ax^5 +c = 0 can have 5 roots real or complex.
So and c cannot be zero. b could be zero under the given condition.
We’ve answered 318,911 questions. We can answer yours, too.Ask a question