The number of distinct real roots of the equa | vedclass

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(A) We analyze the equation $|x||x+2|-5|x+1|-1=0$ by considering different intervals for $x$:Case $1$: $x \geq 0$The equation becomes $x(x+2)-5(x+1)-1

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$3$

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(A) We analyze the equation $|x||x+2|-5|x+1|-1=0$ by considering different intervals for $x$:Case $1$: $x \geq 0$The equation becomes $x(x+2)-5(x+1)-1=0 \implies x^2+2x-5x-5-1=0 \implies x^2-3x-6=0$.The roots are $x = \frac{3 \pm \sqrt{9+24}}{2} = \frac{3 \pm \sqrt{33}}{2}$. Since $x \geq 0$,we take $x = \frac{3+\sqrt{33}}{2}$. (One root)Case $2$: $-1 \leq x The equation becomes $(-x)(x+2)-5(x+1)-1=0 \implies -x^2-2x-5x-5-1=0 \implies x^2+7x+6=0$.$(x+6)(x+1)=0 \implies x=-6, -1$. In the range $[-1, 0)$,only $x=-1$ is valid. (One root)Case $3$: $-2 \leq x The equation becomes $(-x)(x+2)-5(-(x+1))-1=0 \implies -x^2-2x+5x+5-1=0 \implies x^2-3x-4=0$.$(x-4)(x+1)=0 \implies x=4, -1$. Neither value lies in the interval $[-2, -1)$. (No root)Case $4$: $x The equation becomes $(-x)(-(x+2))-5(-(x+1))-1=0 \implies x^2+2x+5x+5-1=0 \implies x^2+7x+4=0$.The roots are $x = \frac{-7 \pm \sqrt{49-16}}{2} = \frac{-7 \pm \sqrt{33}}{2}$.Both $\frac{-7+\sqrt{33}}{2} \approx -0.63$ and $\frac{-7-\sqrt{33}}{2} \approx -6.37$ are candidates. Since $x Total distinct real roots: $1+1+0+1 = 3$.

$f_1(x) = 5 x^2 + 2 x + 1$	$f_2(x) = 5 x^2 + 6 x + 1$
$f_3(x) = x^2 - 7 x + 6$	$f_4(x) = 64 x^2 + 48 x + 9$

The number of distinct real roots of the equation $|x||x+2|-5|x+1|-1=0$ is:

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