Solutions of the equation $x | x + 1 | + 1 = 0$ are

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Solution

$x | x + 1 | + 1 = 0$

if $x > - 1$

$x (x + 1) + 1 = 0$

$\Rightarrow$ $x^{2} + x + 1 = 0$

$x = \frac{- 1 \pm \sqrt{1 - 4}}{2}$

$= \frac{- 1 \pm \sqrt{- 3}}{2}$

$= \frac{- 1 \pm i \sqrt{3}}{2}$

if $x < - 1$

$- x (x + 1) + 1 = 0$

$\Rightarrow$ $x^{2} + x - 1 = 0$

$x = \frac{- 1 \pm \sqrt{5}}{2}$

$= \frac{1}{2} (- 1 \pm \sqrt{5})$

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Similar questions

x^{2}

x | x + 1 | + 1 = 0

\frac{1}{2} (1 \pm \sqrt{5})

\frac{- 1}{2} (- 1 \pm \sqrt{5})

\frac{1}{2} (- 1 + \sqrt{5})

y = f_{1} (x)

y = f_{2} (x)

y d x + d y = - e^{x} y^{2} d y

f_{1} (0) = 1

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f_{1} (x) . f_{2} (x) + x^{2} = 0

(x, y)

1 + x^{2} + 2 x sin ({cos}^{- 1} y) = 0,

{tan}^{- 1} (\frac{1 - x}{1 + x}) = \frac{1}{2} {tan}^{- 1} x, (x > 0)

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