Question

if f(5+x)=f(5-x) for every real x and f(x)=0 has four distinct real roots ,then the sum of roots is: ​

Answer 1

Given : f(5+x)=f(5-x)    for every real x and f(x)=0 has four distinct real roots  

To Find : n the sum of roots

Solution:

f(5+x)=f(5-x)  

Substitute x = x - 5

=> f(5 + x - 5) = f( 5 - ( x - 5))

=> f(x) =  f(10 - x)

Let say m   & n are two roots of  f(x)

Then f(m) = 0  = f(10 - m)  

        f(n) = 0  = f(10 - n)  

f(10 - m)    = 0  => 10 - m is a root

f(10 - n)   = 0 => 10 - n is a root

Then 4 roots are  m  , n , 10 - m & 10 - n

Sum of roots  = m + 10 - m + n + 10 - n

20

the sum of roots is: ​ 20

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Answer 2

Question:

if f(5+x)=f(5-x) for every real x and f(x)=0 has four distinct real roots ,then the sum of roots is:

Solution:

f(5 + x) = f(5 - x)

Put x= x-5

We get, f(x) = f(10 - x)

So, if 'a' is one of the roots, '10-a' must be the root of f(x). Similarly, if 'b' is one of the roots, '10-b' must be the root of f(x).

Sum of the roots = a+(10-a)+b+(10-b) = 20