Real Roots of f(5-x)=f(5+x)

$f(5-x)=f(5+x)$ means the function is symmetrical w.r.t. $x=5$ . If $5-x$ is a root for $x\neq 0$ , then $5+x$ must also be a root, meaning that the roots which are not 5 would pair up. Since $f$ has 5 distinct roots, $5$ must also be a root. The four roots are symmetrical w.r.t. x=5, so $x_1+x_2=x_3+x_4=5*2=10$ . Thus, the sum of all 5 roots is $10+10+5=25$ .

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

let 5-x=y.

then, x=5-y

so, 5+x =5+5-y

=10-y

So f(y)=f(10-y) that is,

f(x)=f(10-x)

Now, f(x) has five distinct real roots. if k is a root of the function, then 10-k is also a root,as we can see from the above result. So the roots occur in pairs. But the roots are 5 in number. So one pair must have both the terms equal to each other. Let the pair be c and 10-c. Then, c=10-c

c=5

So the roots are in the following form: (k,10-k),(m,10-m),(5)

So the sum of the roots is k+10-k+m+10-m+5 =25

they gave 5 distinct roots therefore 5 is a solution because for every other solution there will be another solution so 5 should be a solution for odd number of solutions. now they told that there are 5 solutions and sum of two solutions will always be 5-x+5+x =10 so sum of roots is 10+10+5=25

If $(5+x)$ is a root, then so is $(5-x)$ , hence the roots come in pairs, apart from the case $x=0$ . Since there are an odd number of real roots, one of them must be $5$ , and the other $4$ roots pair up.

Hence, the sum of the roots is $5 + (5 - r_1) + (5 + r_1) + (5 - r_{2}) + (5 + r_{2}) = 25$ .

f(5-x)=f(5+x) is equivalent to f(x)=f(10-x) take f(x)=(x-a)(x-b)(x-c)(x-d)(x-e) where a,b,c,d,e are the roots. as f(x)=0 and f(10-x)=0 are the same thing, we take (x-a)(x-b)(x-c)(x-d)(x-e)=0 and (10-x-a)(10-x-b)(10-x-c)(10-x-d)(10-x-e)=0 for an equation of degree n, we get -(sum of all roots)=co-efficient of x^(n-1) we equate the co-efficients of x^4 and get -(a+b+c+d+e)=(a-10+b-10+c-10+d-10+e-10) thus we get a+b+c+d+e=25

Let y= 5-x. So, x= 5-y So, 5+x= 10-y

If y is a root, so must be 10-y.

So sum of 4 roots= 20 Sum of 5 roots= 25

Let $g(x) = f(x-5)$

Thus, $g(x) = g(-x)$ . This implies that g is even.

Note that a root in g(x) corresponds with a root in f(x)

Note that x=0 is a root, otherwise there would be an even number of roots, because for every root x, we can pair it up with root -x.

Let a and b be 2 roots of g(x) so that $a \neq \pm b$ . Then we also have the 2 roots -a and -b.

This makes 5 roots in total. In fact, this is the only configuration of 5 roots of g(x), a, b, 0, -a and -b.

Thus, the 5 roots of f(x) are a+5, b+5, 5, 5-a and 5-b

Thus the sum of them is 25.