Solve |x^2 – 2x| + |x - 4| > |x^2 – 3x + 4| . Please solve this inequality. I’m not sure how to account for quadratic equations inside modulus.

Thus we write the whole eqn. as:
|x(x – 2)| + |x - 4| > x^2 – 3x + 4
Now define x for modulus,

case 1. when x

after solving,

0 > 0 

which is not possible. hence here xbelongs to null .

 

Similarly for case   4 > x = > 2

0 > 0

x belongs to null.

 

For cases,  0

x (x-2)

x belongs to (0, 2)

 

for case, x > = 4

x belongs to [4 ,infinity)

 

Final answer,

x belongs (0,2) U [4, infinity)

We are given |x²-2x| +|4-x| > |(x²-2x)+(4-x)|

 

From the triangle inequality we know that |a| +|b| >= |a+b| . So we need to eliminate the equality case which happens when a,b, and hence a+b are of the same sign. Note that x²-3x+4 >0 always. So we need to eliminate the case when

 

x²-2x>0 and 4-x>0 i.e when (x-1)²>1 and x

 

Hence the required set is (0,2) U [4, infinity)