The number of distinct real roots of the equation | x | | x + 2 | - 5 | x + 1 | - 1 = 0 is _________ [2024]

Question

The number of distinct real roots of the equation $| x | | x + 2 | - 5 | x + 1 | - 1 = 0$ is _________ [2024]

Smart Achievers · Accepted Answer

(3)

$f (x) = | x | | x + 2 | - 5 | x + 1 | - 1 = 0$

Case 1 : $x \geq 0$

$∴$ $f (x) = x (x + 2) - 5 (x + 1) - 1 = 0$

$\Rightarrow x^{2} + 2 x - 5 x - 6 = 0 \Rightarrow x^{2} - 3 x - 6 = 0$

$\Rightarrow x = \frac{3 \pm \sqrt{9 + 24}}{2}$

$\Rightarrow x = \frac{3 \pm \sqrt{33}}{2},$ One positive root [ $∵$ $x \geq 0$ ]

Case 2: $- 1 \leq x < 0$

$\Rightarrow f (x) = - x (x + 2) - 5 (x + 1) - 1 = 0$

$\Rightarrow - x^{2} - 7 x - 6 = 0 \Rightarrow x^{2} + 7 x + 6 = 0$

$\Rightarrow (x + 6) (x + 1) = 0$

$\Rightarrow x = - 6, - 1$ One root i.e., $x = - 1$

Case 3 : $- 2 \leq x < - 1$

$f (x) = - x (x + 2) + 5 (x + 1) - 1 = 0$

$\Rightarrow - x^{2} - 2 x + 5 x + 4 = 0$

$\Rightarrow - x^{2} + 3 x + 4 = 0 \Rightarrow x^{2} - 3 x - 4 = 0$

$\Rightarrow (x + 1) (x - 4) = 0 \Rightarrow x = - 1, 4$

No root possible in given range of $x$ .

Case 4 : $x < - 2$

$f (x) = - x (- x - 2) + 5 (x + 1) - 1 = 0$

$\Rightarrow x^{2} + 2 x + 5 x + 4 = 0$

$\Rightarrow x^{2} + 7 x + 4 = 0$

$\Rightarrow x = \frac{- 7 \pm \sqrt{49 - 16}}{2} = \frac{- 7 \pm \sqrt{33}}{2}$

One root in the range.

$∴$ We have 3 distinct real roots

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