Question

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

Answer 1

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.