Let f(x)= (x-p)(x-q). If f(11)=f(20)=0, then f(10)=?
A. 1
B. -1
C. 5
D. -5
E. 10
Hi Max@Math Revolution,
Let's take a look at your question.
$$f\left(x\right)=\left(x-p\right)\left(x-q\right)$$
$$f\left(11\right)=\left(11-p\right)\left(11-q\right)$$
$$\text{Since,} f\left(11\right)=0$$
$$\left(11-p\right)\left(11-q\right)=0$$
$$\text{Either }\left(11-p\right)=0 \text{or }\left(11-q\right)=0$$
$$\text{Either }p=11 \text{or }q=11$$
Also ,
$$f\left(20\right)=\left(20-p\right)\left(20-q\right)$$
$$\text{Since,} f\left(20\right)=0$$
$$\left(20-p\right)\left(20-q\right)=0$$
$$\text{Either }\left(20-p\right)=0 \text{or }\left(20-q\right)=0$$
$$\text{Either }p=20 \text{or }q=20$$
Now how we are going to decide which what values of p and q will be selected to find f(10).
We know that f(11)=f(20)=0, therefore,
$$\left(11-p\right)\left(11-q\right)=\left(20-p\right)\left(20-q\right)$$
$$121-11p-11q+pq=400-20p-20q+pq$$
$$121-11p-11q=400-20p-20q$$
$$121-11p-11q-400+20p+20q=0$$
$$-279+9p+9q=0$$
$$-31+p+q=0$$
$$p+q=31$$
So we will be selecting the values of p and q such that there sum is 31.
So we can take p=11, q = 20 or p=20, q=11, either case we will get the same result.
Now we will find f(10).
$$f\left(10\right)=\left(10-p\right)\left(10-q\right)$$
Plugin the values of p and q:
$$f\left(10\right)=\left(10-11\right)\left(10-20\right)=\left(-1\right)\left(-10\right)=10$$
Therefore, Option
E is correct.
Hope it helps.
I am available if you'd like any follow up.