What is the answer to |x^2-2x|+|x-3|=4 ?

The answer to |x^2-2x|+|x-3|=4 is x=(3-sqrt(13))/2 \lor x=(1+sqrt(29))/2

Solve for x: |x^2-2x|+|x-3|=4

The solution is x=(3-sqrt(13))/2 \lor x=(1+sqrt(29))/2

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x^{2} - 2 x + ∣ x - 3 ∣ = 4

x = \frac{3 - 13}{2} or x = \frac{1 + 29}{2}

Solution steps

x^{2} - 2 x + ∣ x - 3 ∣ = 4

Find positive and negative intervals

x \leq 0, 0 < x < 2, 2 \leq x < 3, x \geq 3

Solve the inequality for each interval

x = \frac{3 - 13}{2} or No Solution or x = \frac{1 + 29}{2}

x = \frac{3 - 13}{2} or x = \frac{1 + 29}{2}

2 ∣ 3 - x ∣ + 3 ∣ x - 1 ∣ = 4

∣ x ∣ - 2 x + 5 = 0

[∣ x - 1 ∣] + [∣ 7 - x ∣] = 6

∣ x + 2 ∣ = ∣ x ∣

x^{2} - 6 x + 7 = \frac{( 5 x - 9 )}{3}

What is the answer to |x^2-2x|+|x-3|=4 ?
The answer to |x^2-2x|+|x-3|=4 is x=(3-sqrt(13))/2 \lor x=(1+sqrt(29))/2
Solve for x: |x^2-2x|+|x-3|=4
The solution is x=(3-sqrt(13))/2 \lor x=(1+sqrt(29))/2