|x2-4x+3|<3 Valid proof for x?

Question

First, we should solve the inequality |x2-4x+3|<3 by first adding 4x and subtracting 3 from each side of the inequality giving us |x2|<4x. Next, divide x from each side and we see that |x|< 4 for the inequality to be true. Valid?

Answer 1

-3 which is equivalenet to
0 and
x^2-4x<0......(2)

Solve for (1)
x can be any real number

Solve for (2)
0
So the final solution is
0

Answer 2

Let x = 0

|0² - 4*0 +3|<3
|0 - 0 + 3| < 3
|3| < 3
±3 < 3

3 < 3 is false so the assertion that it is true for |x| < 4 is false as shown by a counter example.

Answer 3

x^2-4x+3=(x-1)(x-3)

|1^2-4(1)+3|<3
0<3 True

|3^2-4(3)+3|<3
0<3 True

Answer 4

To prove the result, try substituting the answer back into the original equation.

Try 3.

|3^2-4*3+3|
|9 -12 +3|
|0| < 3

If you have a calculator try 3.5 and 4.

Answer 5

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Answer 6

This can be re-written as

x^2 - 4x <0

x(x-4) < 0 x=0 or x-4=0

therefore 0
What you have written is only half the answer.