CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Let f(x)=x2px+q, p is an odd positive integer and the roots of the equation f(x)=0 are two distinct prime numbers, if p+q=35, then the value of f(10)(10r=1f(r))878 equal to  ___


Solution

Roots of x2px+q=0 are two distinct prime numbers.
Sum of roots = p = odd positive integer (given)
 Both roots of the equation cannot be odd.
 One root must be 2(prime number)
Then (2)2p(2)+q=0
or 2pq=4
and given p+q=35
Solving Eqn. (i) and (ii), we get
p=13,q=22
f(x)=x213x+22
10r=1r21310r=1r+2210r=11
=10.11.21613.10.112+22.10
=110 and f(10)=8
f(10)10r=1f(r)=(8)(100)=880

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
QuestionImage
QuestionImage
View More...



footer-image