To solve the inequality \left|x^2-4x+3\right|\le\left|x-3\right|-x, we need to analyze the expressions on both sides of the inequality separately and then compare them.

Step 1: Analyze \left|x^2 - 4x + 3\right|

The expression inside the absolute value is a quadratic: x^2 - 4x + 3. We can factor it:

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

The roots of this quadratic are x = 1 and x = 3. The sign of the quadratic changes at these points. We can analyze the sign of x^2 - 4x + 3 in the intervals determined by these roots:

For x < 1, both factors (x - 1) and (x - 3) are negative, so the product is positive.
For 1 < x < 3, (x - 1) is positive and (x - 3) is negative, so the product is negative.
For x > 3, both factors are positive, so the product is positive.

Thus, the absolute value \left|x^2 - 4x + 3\right| can be expressed as:

\left|x^2 - 4x + 3\right| = \begin{cases} x^2 - 4x + 3, & x \leq 1 \text{ or } x \geq 3 \\ -(x^2 - 4x + 3), & 1 < x < 3 \end{cases}

Step 2: Analyze \left|x - 3\right| - x

The expression \left|x - 3\right| depends on whether x is less than or greater than 3:

For x < 3, \left|x - 3\right| = 3 - x.
For x \geq 3, \left|x - 3\right| = x - 3.

Thus, \left|x - 3\right| - x can be expressed as:

\left|x - 3\right| - x = \begin{cases} 3 - 2x, & x < 3 \\ -3, & x \geq 3 \end{cases}

Step 3: Solve the Inequality

We now solve the inequality \left|x^2 - 4x + 3\right| \leq \left|x - 3\right| - x by considering the different cases .

Case 1: x \leq 1

In this interval, \left|x^2 - 4x + 3\right| = x^2 - 4x + 3 and \left|x - 3\right| - x = 3 - 2x.

The inequality becomes:

x^2 - 4x + 3 \leq 3 - 2x

Simplifying, we get:

x^2 - 2x \leq 0

Factoring gives:

x(x - 2) \leq 0

The solution to this inequality is 0 \leq x \leq 2. However, since we are considering x \leq 1, the solution in this case is 0 \leq x \leq 1 .

Case 2: 1 < x < 3

In this interval, \left|x^2 - 4x + 3\right| = -(x^2 - 4x + 3) and \left|x - 3\right| - x = 3 - 2x.

The inequality becomes:

-(x^2 - 4x + 3) \leq 3 - 2x

Simplifying, we get:

-x^2 + 4x - 3 \leq 3 - 2x

Rearranging terms gives:

-x^2 + 6x - 6 \leq 0

Multiplying through by -1 (and reversing the inequality sign):

x^2 - 6x + 6 \geq 0

The roots of the equation x^2 - 6x + 6 = 0 are:

x = \frac{6 \pm \sqrt{36 - 24}}{2} = \frac{6 \pm \sqrt{12}}{2} = 3 \pm \sqrt{3}

The inequality x^2 - 6x + 6 \geq 0 holds for x \leq 3 - \sqrt{3} or x \geq 3 + \sqrt{3}. However, since we are considering 1 < x < 3, the solution in this case is 1 < x \leq 3 - \sqrt{3} .

Case 3: x \geq 3

In this interval, \left|x^2 - 4x + 3\right| = x^2 - 4x + 3 and \left|x - 3\right| - x = -3.

The inequality becomes:

x^2 - 4x + 3 \leq -3

Simplifying, we get:

x^2 - 4x + 6 \leq 0

The roots of the equation x^2 - 4x + 6 = 0 are complex, so the quadratic is always positive. Thus, there are no solutions in this interval

Conclusion

Combining cases 1 and 2:

0\le x\le3-\sqrt3

General

Modulus & Inequalities - Practice problems