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

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(C) We analyze the equation $x|x-2|+3|x-3|+1=0$ in three cases:Case $(I): x < 2$The equation becomes $x(2-x) + 3(3-x) + 1 = 0$$-x^2 + 2x + 9 - 3x + 1

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$1$

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(C) We analyze the equation $x|x-2|+3|x-3|+1=0$ in three cases:Case $(I): x The equation becomes $x(2-x) + 3(3-x) + 1 = 0$$-x^2 + 2x + 9 - 3x + 1 = 0$$-x^2 - x + 10 = 0 \Rightarrow x^2 + x - 10 = 0$The roots are $x = \frac{-1 \pm \sqrt{1 - 4(1)(-10)}}{2} = \frac{-1 \pm \sqrt{41}}{2}$.Since $\sqrt{41} \approx 6.4$,$x_1 = \frac{5.4}{2} = 2.7$ (not in range) and $x_2 = \frac{-7.4}{2} = -3.7$ (in range).So,$1$ root here.Case $(II): 2 \leq x The equation becomes $x(x-2) + 3(3-x) + 1 = 0$$x^2 - 2x + 9 - 3x + 1 = 0$$x^2 - 5x + 10 = 0$The discriminant $D = (-5)^2 - 4(1)(10) = 25 - 40 = -15 No real roots in this interval.Case $(III): x \geq 3$The equation becomes $x(x-2) + 3(x-3) + 1 = 0$$x^2 - 2x + 3x - 9 + 1 = 0$$x^2 + x - 8 = 0$The roots are $x = \frac{-1 \pm \sqrt{1 - 4(1)(-8)}}{2} = \frac{-1 \pm \sqrt{33}}{2}$.Since $\sqrt{33} \approx 5.74$,$x_3 = \frac{4.74}{2} = 2.37$ (not in range) and $x_4 = \frac{-6.74}{2} = -3.37$ (not in range).No real roots in this interval.Thus,there is only $1$ real root.

The number of real roots of the equation $x|x-2|+3|x-3|+1=0$ is:

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