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

The answer to |x^2+x-4|=|x^2-4|+|x| is -2 = 2

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|x^2+x-4|=|x^2-4|+|x|

x^{2} + x - 4 = x^{2} - 4 + ∣ x ∣

- 2 \leq x \leq 0 or x \geq 2

Solution steps

x^{2} + x - 4 = x^{2} - 4 + ∣ x ∣

Find positive and negative intervals

x \leq \frac{- 17 - 1}{2}, \frac{- 17 - 1}{2} < x \leq - 2, - 2 < x < 0, 0 \leq x < \frac{17 - 1}{2}, \frac{17 - 1}{2} \leq x < 2, x \geq 2

Solve the inequality for each interval

No Solution or x = - 2 or - 2 < x < 0 or x = 0 or x \geq 2

Merge Overlapping Intervals

- 2 \leq x \leq 0 or x \geq 2

∣ x - 1 ∣ + ∣ 3 x + 6 ∣ = 7

∣ 7 x - 12 ∣ - ∣ 7 x - 11 ∣ = 1

x^{2} - 2 x - 8 = 2 x

∣ x - 2 ∣ + ∣ x - 4 ∣ = 6

∣ x + 3 ∣ - ∣ 2 x - 5 ∣ = 7

What is the answer to |x^2+x-4|=|x^2-4|+|x| ?
The answer to |x^2+x-4|=|x^2-4|+|x| is -2<= x<= 0\lor x>= 2
Solve for x: |x^2+x-4|=|x^2-4|+|x|
The solution is -2<= x<= 0\lor x>= 2