M61-03

Question

Let  f(x)= (x-p)(x-q) . If  f(11)=f(20)=0 , then  f(10)= ?

A. -5
B. -1
C. 1
D. 5
E. 10

MathRevolution · Answer

Official Solution:

Let  f(x)= (x-p)(x-q) . If  f(11)=f(20)=0 , then  f(10)= ?

A. -5
B. -1
C. 1
D. 5
E. 10

Since 11 and 20 are roots of  f(x) = 0 , we must have  f(x) = (x-11)(x-20) . Thus,  f(10) = (10-11)(10-20) = (-1)(-10) = 10 .

Answer: E

baru · Answer

Hi Bunuel ,

f(x)=(x−p)(x−q)

f(11)=f(20)

If I take p and q as 15.5 then also f(11) = F(20) , I was not given that p & Q are root .(x−p)(x−q)

F(11)=(11-15.5)(11.15.5) = F(20) = (20-11.5)(20-11.5)

Please correct me , I am confused .

MathRevolution · Answer

You didn't consider f(11) = 0 and f(20) = 0.

Zain90 · Answer

I think this the explanation isn't clear enough, please elaborate. I thought 11 and 20 will replace x in the F(x) and I got answer 1. I equated f(11) to f(20) to get values in terms of p+q and pq and then substituted in answer

mimishaf · Answer

You can solve it by using the quadratic equation method which is somewhat longer but might help you understand.

Using the information above.

f(11)=f(20)

Thus:
(11-p)(11-q) = (20-p)(20-q)
Solving both sides of the question:

11^2 -11p - 11q + pq = 20^2 - 20p - 20q + pq
9p + 9q = 20^2 - 11^2
9 (p+q) = (20+11) (20-11)

p + q = 20 + 11

So in effect you get the values for both P and Q, which for the purpose of this question are reversible. 
Therefore:

f(10) = (10 - 20)(10 - 11)
f(10) =  10

LeoN88 · Answer

The product of two numbers is zero when either of/both the numbers is zero.
So we can assume: (p,q) : (11,20) in any order.

So f(10)= (10-11)(10-20)
answer 10.

ritwikk2 · Answer

If we take value of both p and q as 11

we get f(x) = (x-11)(x-11)
so f(10) = (10-11)(10-11) = (-1)(-1) = 1

Choice (C).

can someone explain what is the issue in my approach.

gurmukh · Answer

If you take both p and q as 11 then f(20) cannot be zero hence p and q both 11 is wrong.

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