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

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

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

The solution is x=-2(1+sqrt(2))\lor x=-2\lor x=2(sqrt(2)-1)

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x^{2} + 4 x = 4

x = - 2 (1 + 2) or x = - 2 or x = 2 (2 - 1)

Decimal Notation

x = - 4.82842 \dots, x = - 2, x = 0.82842 \dots

Solution steps

x^{2} + 4 x = 4

Apply absolute rule: If

∣ u ∣ = a, a > 0

then

u = a

u = - a

x^{2} + 4 x = - 4 or x^{2} + 4 x = 4

x^{2} + 4 x = - 4 or x^{2} + 4 x = 4

x = - 2 or (x = - 2 (1 + 2) or x = 2 (2 - 1))

x = - 2 (1 + 2) or x = - 2 or x = 2 (2 - 1)

10 b^{3} + 15 b^{2}

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

p + 17.5 = 17.8

612

\frac{1}{3} y \geq 5

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