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

The solution to |x^2-4x+3|>x is x (sqrt(13)+5)/2

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|x^2-4x+3|>x

x^{2} - 4 x + 3 > x

x < \frac{- 13 + 5}{2} or x > \frac{13 + 5}{2}

Interval Notation

(- \infty, \frac{- 13 + 5}{2}) \cup (\frac{13 + 5}{2}, \infty)

Decimal Notation

x < 0.69722 \dots or x > 4.30277 \dots

Solution steps

x^{2} - 4 x + 3 > x

Find positive and negative intervals

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

Solve the inequality for each interval

x < \frac{- 13 + 5}{2} or No Solution or x > \frac{13 + 5}{2}

Merge Overlapping Intervals

x < \frac{- 13 + 5}{2} or x > \frac{13 + 5}{2}

472

x + 2 \geq 0

2 x - 3 + 2 x = 7 - x

e^{- 2 l n (9)}

m^{\frac{1}{3}} \cdot m^{\frac{1}{5}}

What is |x^2-4x+3|>x ?
The solution to |x^2-4x+3|>x is x<(-sqrt(13)+5)/2 \lor x>(sqrt(13)+5)/2