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

Question

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

Accepted Answer

(I) $x < 2$: $|x-2|=2-x$, $|x-3|=3-x$.
$$x(2-x)+3(3-x)+1=0 \implies -x^2-x+10=0 \implies x^2+x-10=0 \implies x=\frac{-1\pm\sqrt{41}}{2}$$
Since $\sqrt{41}\approx6.4$, only $x=\dfrac{-1-\sqrt{41}}{2}\approx-3.7$ satisfies $x<2$.

(II) $2 \le x < 3$: $|x-2|=x-2$, $|x-3|=3-x$.
$$x(x-2)+3(3-x)+1=0 \implies x^2-5x+10=0,\quad D=25-40<0$$
No real roots.

(III) $x \ge 3$: $|x-2|=x-2$, $|x-3|=x-3$.
$$x(x-2)+3(x-3)+1=0 \implies x^2+x-8=0 \implies x=\frac{-1\pm\sqrt{33}}{2}$$
Since $\sqrt{33}\approx5.74$, neither root ($\approx2.37$ or $\approx-3.37$) satisfies $x\ge3$.

Total: only 1 real root, $x=\dfrac{-1-\sqrt{41}}{2}$.

Solutions · 1