The answer I got for the problem you gave me is: 2<x<3 or in interval notation: (2,3)
You are correct, but you still are not showing your work. Just telling you whether your answer is right or wrong does not let us see what you are thinking. So let's walk through this example.
\(\displaystyle 4x^2 - 20x + 25 = 1 \implies 4x^2 - 20x + 24 = 0 \implies 4(x^2 - 5x + 6) = 0 \implies \\
4(x - 2)(x - 3) = 0 \implies x = 2 \text { or } x = 3.\)
Now why did I suggest starting with the equality? Because the function
\(\displaystyle f(x) = 4x^2 - 20x + 25\)
is continuous, it will be greater or less than a given number EVERYWHERE in domains to one side of a point of intersection. It will equal that number only at points of intersection. We have 2 points of intersection so we have 2 + 1 = 3 domains to test. So in the domain less than 2, pick a convenient value like x = 0.
\(\displaystyle 4(0^2) - 20(0) + 25 = 25 > 1 \implies f(x) > 1 \text { if } x < 2.\)
And we know that \(\displaystyle f(x) = 1 \text { if } x = 2.\)
Now pick a convenient value for x in the domain between 2 and 3 such as 2.5.
\(\displaystyle 4(2.5^2) - 20(2.5) + 25 = 4(6.25) - 50 + 25 = 25 - 25 = 0 < 1 \implies\\
f(x) < 1 \text { if } 2 < x < 3.\)
And we know that \(\displaystyle f(x) = 1 \text { if } x = 3.\)
Now pick a convenient value in the domain greater than 3 like 10.
\(\displaystyle 4(10^2) - 20(10) + 25 = 400 - 200 + 25 = 225 > 1 \implies f(x) > 1 \text { if } x > 3.\)
You now know the relationship between f(x) and 1 for every value of x and so can answer any question about inequalities between f(x) and 1.
Try this one. What is the domain where
\(\displaystyle 4x^2 - 20x + 25 < -\ 3.\)