What is |x^2-4x+3|>= |x^2-6x+8| ?

The solution to |x^2-4x+3|>= |x^2-6x+8| is (-sqrt(3)+5)/2 = (sqrt(3)+5)/2

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|x^2-4x+3|>= |x^2-6x+8|

x^{2} - 4 x + 3 \geq x^{2} - 6 x + 8

\frac{- 3 + 5}{2} \leq x \leq \frac{5}{2} or x \geq \frac{3 + 5}{2}

Solution steps

x^{2} - 4 x + 3 \geq x^{2} - 6 x + 8

Find positive and negative intervals

x \leq 1, 1 < x \leq 2, 2 < x < 3, 3 \leq x < 4, x \geq 4

Solve the inequality for each interval

No Solution or \frac{- 3 + 5}{2} \leq x \leq 2 or 2 < x \leq \frac{5}{2} or \frac{3 + 5}{2} \leq x < 4 or x \geq 4

Merge Overlapping Intervals

\frac{- 3 + 5}{2} \leq x \leq \frac{5}{2} or x \geq \frac{3 + 5}{2}

\frac{∣ x - 5 ∣}{∣ x + 2 ∣} < 4

x^{2} - 3 x + 2 > 2

x^{2} - 3 x + 2 > 0

∣ 3 x - 2 ∣ > 3 ∣ x - 1 ∣

∣ x - 1 ∣ + ∣ x - 2 ∣ \leq 3

What is |x^2-4x+3|>= |x^2-6x+8| ?
The solution to |x^2-4x+3|>= |x^2-6x+8| is (-sqrt(3)+5)/2 <= x<= 5/2 \lor x>= (sqrt(3)+5)/2