The number of real roots of the equation x |x | vedclass

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(B) We analyze the equation $x|x| - 5|x + 2| + 6 = 0$ by considering different intervals for $x$.Case $1$: $x \ge 0$.The equation becomes $x^2 - 5(x +

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$3$

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(B) We analyze the equation $x|x| - 5|x + 2| + 6 = 0$ by considering different intervals for $x$.Case $1$: $x \ge 0$.The equation becomes $x^2 - 5(x + 2) + 6 = 0$,which simplifies to $x^2 - 5x - 4 = 0$.The roots are $x = \frac{5 \pm \sqrt{25 - 4(1)(-4)}}{2} = \frac{5 \pm \sqrt{41}}{2}$.Since $x \ge 0$,we accept $x = \frac{5 + \sqrt{41}}{2}$. ($1$ root)Case $2$: $-2 \le x The equation becomes $-x^2 - 5(x + 2) + 6 = 0$,which simplifies to $-x^2 - 5x - 4 = 0$,or $x^2 + 5x + 4 = 0$.Factoring gives $(x + 1)(x + 4) = 0$,so $x = -1$ or $x = -4$.Since $-2 \le x Case $3$: $x The equation becomes $-x^2 - 5(-(x + 2)) + 6 = 0$,which simplifies to $-x^2 + 5x + 10 + 6 = 0$,or $x^2 - 5x - 16 = 0$.The roots are $x = \frac{5 \pm \sqrt{25 - 4(1)(-16)}}{2} = \frac{5 \pm \sqrt{89}}{2}$.Since $x Total number of real roots is $1 + 1 + 1 = 3$.

The number of real roots of the equation $x |x| - 5|x + 2| + 6 = 0$ is:

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