question is good one…

Here, we have to take two condition-

After removing modulus we can write equation as follow-

+(x^2-4x)<5 …….. (1)

And

-(x^2-4x)<5 ……… (2)

Case-1_

Solving inequation ..(1)

+(x^2-4x)<5

Take 5 to the LHS -

(x^2-4x)-5 < 0

After solving LHS[which is quadratic]-

We get_

(x+1)(x-5)<0

=> x belongs to (-1 to 5)……. (3)

Case 2_

Solving inequation….(2)

-(x^2-4x)<5

Take LHS term to RHS -

We get -

0 < (x^2-4x)+5

But equation in RHS gives imaginary roots .That's why we have to exclude the second case .

Therefore from …(3) , we get -

Answer-> -1 < x < 5

f(x) = x^2 - 4x - 5 < 0

First, find the 2 real roots: f(x) = 0.Since a -b + c = 0, one real root is (-1) and the other is (-c/a = 5)

------------------|-1===|0===========|5-----------------

Test point method. Plot (-1) and (5) on the number line. Use the origin O as test point --> x = 0 --> f (x) = -5 < 0. True. Then the origin O is located on the segment (-1, 5) that is the solution set.Answer: Open interval (-1, 5).

Algebraic Method. Between the 2 real roots (-1) and (5), f(x) is negative, having the opposite sign to a (> 0).Answer: Open interval (-1, 5)