Question

The equation $||x-a|-b|=c$ has four distinct real roots, then

• A. $a>b-c>0$
• B. $c>b>0$
• C. $a>c+b>0$
• D. $b>c>0$

Answer 1

Answer: (D)

b>c>0

Answer 2

The given equation is $||x-a|-b|=c$. For this equation to have real roots, $c$ must be non-negative. If $c=0$, the equation becomes $||x-a|-b|=0$, which implies $|x-a|-b=0$, or $|x-a|=b$. For $|x-a|=b$ to have real roots, $b \ge 0$. If $b=0$, $|x-a|=0$, which gives $x=a$ (one root). If $b>0$, $|x-a|=b$ gives $x-a=\pm b$, so $x=a \pm b$ (two distinct roots). Thus, if $c=0$, we can have at most two distinct roots. For four distinct roots, we must have $c>0$.

With $c>0$, the equation $||x-a|-b|=c$ splits into two equations:

1. $|x-a|-b = c \implies |x-a| = b+c$
2. $|x-a|-b = -c \implies |x-a| = b-c$

For equation (1), $|x-a|=b+c$, to have real roots, we must have $b+c \ge 0$. Since $c>0$, this condition is satisfied if $b \ge -c$. If $b+c > 0$, this equation gives two distinct roots: $x-a = \pm (b+c)$, so $x = a \pm (b+c)$. If $b+c = 0$, this equation gives one root: $x-a = 0$, so $x=a$. If $b+c < 0$, this equation has no real roots.

For equation (2), $|x-a|=b-c$, to have real roots, we must have $b-c \ge 0$. If $b-c > 0$, this equation gives two distinct roots: $x-a = \pm (b-c)$, so $x = a \pm (b-c)$. If $b-c = 0$, this equation gives one root: $x-a = 0$, so $x=a$. If $b-c < 0$, this equation has no real roots.

For the original equation to have four distinct real roots, we need both equations (1) and (2) to contribute roots, and the total number of distinct roots must be four.

Case I: Equation (1) gives two distinct roots and Equation (2) gives two distinct roots. This requires $b+c > 0$ and $b-c > 0$. $b-c > 0 \implies b > c$. Since we assumed $c>0$, $b>c$ implies $b>0$. If $b>c>0$, then $b+c > c > 0$, so $b+c>0$ is automatically satisfied. So, if $b>c>0$, Equation (1) gives two distinct roots $a+b+c$ and $a-b-c$, and Equation (2) gives two distinct roots $a+b-c$ and $a-b+c$. The four potential roots are $a+b+c$, $a-b-c$, $a+b-c$, $a-b+c$. We need to check if these four roots are distinct. Since $b>c>0$, we have $b+c > b-c > 0$. The values $b+c$, $-(b+c)$, $b-c$, $-(b-c)$ are all distinct. $b+c \ne b-c$ (since $c \ne 0$). $b+c \ne -(b-c)$ (since $b \ne 0$). $-(b+c) \ne b-c$ (since $b \ne 0$). $-(b+c) \ne -(b-c)$ (since $c \ne 0$). Since the four values $+(b+c), -(b+c), +(b-c), -(b-c)$ are distinct, the four roots $a+(b+c), a-(b+c), a+(b-c), a-(b-c)$ are distinct. Thus, the condition $b>c>0$ guarantees four distinct real roots.

Case II: One equation gives two distinct roots and the other gives one root. This would require either ($b+c>0$ and $b-c=0$) or ($b+c=0$ and $b-c>0$). If $b-c=0$, then $b=c$. Since $c>0$, $b=c>0$. Then $b+c = c+c = 2c > 0$. Equation (1) is $|x-a|=2c$, which gives two distinct roots $a \pm 2c$. Equation (2) is $|x-a|=0$, which gives one root $a$. The roots are $a-2c, a, a+2c$. These are three distinct roots. So $b=c>0$ gives three distinct roots. If $b+c=0$, since $c>0$, this implies $b=-c<0$. Then $b-c = -c-c = -2c < 0$. Equation (1) is $|x-a|=0$, which gives one root $a$. Equation (2) is $|x-a|=-2c$, which has no real roots since $-2c<0$. In this case, there is only one distinct root.

Case III: Both equations give one root. This requires $b+c=0$ and $b-c=0$. This implies $2b=0 \implies b=0$ and $2c=0 \implies c=0$. But we require $c>0$ for four distinct roots. So this case is not possible.

Case IV: One equation gives two distinct roots and the other gives no roots. This requires either ($b+c>0$ and $b-c<0$) or ($b+c<0$ and $b-c>0$). If $b+c>0$ and $b-c<0$, then $b<c$. Since $c>0$, $b$ can be positive, zero or negative. If $b<c$ and $c>0$, then $b+c$ could be positive (e.g., $b=1, c=2$), zero (e.g., $b=-2, c=2$), or negative (e.g., $b=-3, c=2$). We need $b+c > 0$. So we need $b<c$ and $b+c>0$. This implies $-c < b < c$. If $-c < b < c$, then $b+c > 0$ and $b-c < 0$. Equation (1) $|x-a|=b+c$ gives two distinct roots $a \pm (b+c)$. Equation (2) $|x-a|=b-c$ has no real roots. In this case, there are only two distinct real roots. If $b+c<0$ and $b-c>0$, then $b<-c$ and $b>c$. This is impossible since $c>0$ implies $-c<c$.

Based on the analysis, the condition for four distinct real roots is $b>c>0$. Let's check the given options: (A) $a>b-c>0$. This implies $b-c>0$, so $b>c$. The condition $a>b-c$ is irrelevant to the number of roots. For example, if $a=1$, $b=5$, $c=2$, then $b>c>0$ holds, but $a \not> b-c$ ($1 \not> 3$). The equation has four distinct roots. So (A) is not the correct condition. (B) $c>b>0$. This implies $b-c<0$. As shown in Case IV (with $b>0$), this gives only two distinct roots. So (B) is incorrect. (C) $a>c+b>0$. This implies $c+b>0$. The condition $a>c+b$ is irrelevant. If $b>c>0$, then $c+b>0$. But $b>c>0$ is the correct condition. For example, if $a=1$, $b=5$, $c=2$, then $b>c>0$ holds, but $a \not> c+b$ ($1 \not> 7$). So (C) is not the correct condition. (D) $b>c>0$. As derived, this condition guarantees four distinct real roots.

The final answer is $\boxed{b>c>0}$.

The final answer is $\boxed{b>c>0}$.