The number of real solutions of the equation | vedclass

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(D) We analyze the equation $x|x+5|+2|x+7|-2=0$ by considering three cases based on the critical points $x = -5$ and $x = -7$.Case $I$: $x \geq -5$The

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$3$

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(D) We analyze the equation $x|x+5|+2|x+7|-2=0$ by considering three cases based on the critical points $x = -5$ and $x = -7$.Case $I$: $x \geq -5$The equation becomes $x(x+5) + 2(x+7) - 2 = 0$.$x^2 + 5x + 2x + 14 - 2 = 0$$x^2 + 7x + 12 = 0$$(x+3)(x+4) = 0$$x = -3$ or $x = -4$. Both satisfy $x \geq -5$.Case $II$: $-7 The equation becomes $x(-(x+5)) + 2(x+7) - 2 = 0$.$-x^2 - 5x + 2x + 14 - 2 = 0$$-x^2 - 3x + 12 = 0$$x^2 + 3x - 12 = 0$Using the quadratic formula,$x = \frac{-3 \pm \sqrt{9 - 4(1)(-12)}}{2} = \frac{-3 \pm \sqrt{57}}{2}$.Since $\sqrt{57} \approx 7.55$,$\frac{-3 - 7.55}{2} = -5.275$ (which is in the interval $(-7, -5)$) and $\frac{-3 + 7.55}{2} = 2.275$ (rejected).Case $III$: $x \leq -7$The equation becomes $x(-(x+5)) + 2(-(x+7)) - 2 = 0$.$-x^2 - 5x - 2x - 14 - 2 = 0$$-x^2 - 7x - 16 = 0$$x^2 + 7x + 16 = 0$.The discriminant $D = 49 - 64 = -15 The real solutions are $x = -3, -4, \frac{-3-\sqrt{57}}{2}$.Thus,the total number of real solutions is $3$.

The number of real solutions of the equation $x|x+5|+2|x+7|-2=0$ is .....................

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