The number of distinct real roots of the equa | vedclass

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(B) Let $f(x) = |x+1||x+3|-4|x+2|+5 = 0$. Let $t = x+2$. Then $x+1 = t-1$ and $x+3 = t+1$.The equation becomes $|t-1||t+1|-4|t|+5 = 0$,which simplifie

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$2$

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(B) Let $f(x) = |x+1||x+3|-4|x+2|+5 = 0$. Let $t = x+2$. Then $x+1 = t-1$ and $x+3 = t+1$.The equation becomes $|t-1||t+1|-4|t|+5 = 0$,which simplifies to $|t^2-1|-4|t|+5 = 0$.Case $1$: $|t| \geq 1$ $(t^2 \geq 1)$$t^2-1-4|t|+5 = 0 \implies |t|^2-4|t|+4 = 0 \implies (|t|-2)^2 = 0 \implies |t|=2$.So $t=2$ or $t=-2$. Since $x=t-2$,$x=0$ or $x=-4$.Case $2$: $|t| $1-t^2-4|t|+5 = 0 \implies t^2+4|t|-6 = 0$.Let $u = |t|$,then $u^2+4u-6=0$. $u = \frac{-4 \pm \sqrt{16+24}}{2} = -2 \pm \sqrt{10}$.Since $u = |t| \geq 0$,$u = \sqrt{10}-2 \approx 3.16-2 = 1.16$.But we assumed $u The distinct real roots are $x=0$ and $x=-4$. The number of distinct real roots is $2$.

The number of distinct real roots of the equation $|x+1||x+3|-4|x+2|+5=0$ is ...........

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