Solution of the equation $x |x + 1| + 1 = 0$

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Solution

Given $x |x + 1| + 1 = 0$
We know that $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ where $a, b, c$ are constant.
Step 1: Solve the equation for $x > - 1$
if $x > - 1$ , then $x |x + 1| + 1 = 0$
$\begin{array}{rcl} \Rightarrow & x (x + 1) + 1 = 0 \\ x^{2} + x + 1 & = & 0 \\ a & = & 1 b = 1 c = 1 \\ ∴ x & = & \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ x & = & \frac{- 1 \pm \sqrt{1^{2} - 4 \times 1 \times 1}}{2 \times 1} \\ x & = & \frac{- 1 \pm \sqrt{1 - 4}}{2} \\ x & = & \frac{- 1 \pm \sqrt{- 3}}{2} \end{array}$
Step 2: Solve the equation for $x < - 1$
if $x < - 1$ , then $x |x + 1| + 1 = 0$
$\begin{array}{rcl} \Rightarrow & - x (x + 1) + 1 = 0 \\ - x^{2} - x + 1 & = & 0 \\ a & = & - 1 b = - 1 c = 1 \\ ∴ x & = & \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ x & = & \frac{- (- 1) \pm \sqrt{- 1^{2} - 4 \times - 1 \times 1}}{2 \times - 1} \\ x & = & \frac{1 \pm \sqrt{1 + 4}}{- 2} \\ x & = & \frac{- 1 \pm \sqrt{5}}{2} \end{array}$
Hence, $x = \frac{- 1 \pm \sqrt{5}}{2} o r x = \frac{- 1 \pm \sqrt{- 3}}{2}$

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