What is the answer to |x^2-4|=|2x+1| ?

The answer to |x^2-4|=|2x+1| is x=-3\lor x=1-sqrt(6)\lor x=1\lor x=1+sqrt(6)

Solve for x: |x^2-4|=|2x+1|

The solution is x=-3\lor x=1-sqrt(6)\lor x=1\lor x=1+sqrt(6)

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x^{2} - 4 = ∣ 2 x + 1 ∣

x = - 3 or x = 1 - 6 or x = 1 or x = 1 + 6

Decimal

x = - 3, x = - 1.44948 \dots, x = 1, x = 3.44948 \dots

Solution steps

x^{2} - 4 = ∣ 2 x + 1 ∣

Find positive and negative intervals

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

Solve the inequality for each interval

x = - 3 or x = 1 - 6 or x = 1 or x = 1 + 6

∣ 3 v + 6 ∣ = 0

∣ 3 x + 9 ∣ = 10 - x

0 = ∣ x - 4 ∣ - 2

∣ x + 10 ∣ = 7

∣ 4 x - 7 ∣ = 4

What is the answer to |x^2-4|=|2x+1| ?
The answer to |x^2-4|=|2x+1| is x=-3\lor x=1-sqrt(6)\lor x=1\lor x=1+sqrt(6)
Solve for x: |x^2-4|=|2x+1|
The solution is x=-3\lor x=1-sqrt(6)\lor x=1\lor x=1+sqrt(6)