Question

Let $p$ and $q$ be the two distinct solutions to the equation $$(x-3)(x+3) = 21x - 63.$$

If $p > q$, what is the value of $p - q$?

Answer 1

The answer is 5, the "Expert Verified" is wrong :)

Explanation:

First we try factoring the left side to simplify it. Now we can multiply both sides by (x+5)and solve.

Answer 2

Given:

p and q are the two distinct solutions to the equation

`(x-3)(x+3)=21x-63`

To find:

The value of p-q if p>q.

Solution:

We have,

`(x-3)(x+3)=21x-63`

`(x-3)(x+3)=21(x-3)`

`(x-3)(x+3)-21(x-3)=0`

`(x-3)(x+3-21)=0`

`(x-3)(x-18)=0`

Using zero product property, we get

`x-3=0\text{ and }x-18=0`

`x=3\text{ and }x=18`

Here, 18>3, so p=18 and q=3.

Now,

`p-q=18-3`

`p-q=12`

Therefore, the value of p-q is 12.