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:
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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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