There is so much controversy about this question on the internet (well not this question exactly but the idea of taking a reciprocal of a discontinuous function).
Some say you can if you make it clear that the point causing the trouble is excluded in some way. Others say that you cannot contemplate the idea. You are creating a meaningless situation with no definition. It really depends on what you have been told about division by 0. There are ways of getting around this, but you are not taking a calculus course and therefore you likely don't have the tools to deal with it. In any event, it does not look to me like you know about limits yet.
Your marker or teacher or tutor can go either way on this problem and be justified in marking you wrong no matter what you do. As instructed I will put what I think should be done in the comment section. And remember, I'm counting noses and going with the majority when I answer this. It's not anything I'm 100% certain of, but neither is anyone else.
The roots of the quadratic equation 
are
.
Here the given quadratic equation is 
. Here 
.
The roots are
The given ratios 3: 5 and 15: 25 are equal. Because when you divide the ratio 15: 25 by 5 on both numerator and denominator, the first ratio 3: 5 can be obtained. Similarly, when you multiply the first ratio 3: 5 by 5, the ratio 15: 25 can be obtained.
Minus 12x both sides
0=5x^2-12x+9
use quadratic formula
for 0=ax^2+bx+c
x=
given
5x^2-12x+9
a=5
b=-12
c=9
remember: i=√-1
Answer:
Remember that £ has a greater value compared to $ and vice versa.
