Question

If the roots of the equation \(\dfrac{x+a}{x+a+c} + \dfrac{x+b}{x+b+c} = 1\) are equal in magnitude but opposite in sign, then:

• A. \(c \geq a\)
• B. \(a \geq c\)
• C. \(a + b = 0\)
• D. \(a = b\)

Hint:

Clear denominators to get a plain quadratic in x, then use the fact that roots equal in magnitude but opposite in sign must sum to zero.

Answer 1

The Correct Option is C

Step 1: Clear the denominators.
Let \(u=x+a\) and \(v=x+b\), so the denominators become \(u+c\) and \(v+c\). The equation is
\[ \frac{u}{u+c} + \frac{v}{v+c} = 1 \]
Multiply both sides by \((u+c)(v+c)\):
\[ u(v+c) + v(u+c) = (u+c)(v+c) \]

Step 2: Expand both sides.
Left side: \(uv+uc+vu+vc = 2uv+uc+vc\).
Right side: \(uv+uc+vc+c^2\).
Setting them equal:
\[ 2uv+uc+vc = uv+uc+vc+c^2 \]

Step 3: Simplify.
The \(uc\) and \(vc\) terms cancel from both sides, leaving
\[ 2uv = uv + c^2 \implies uv = c^2 \]
Substituting back \(u=x+a\), \(v=x+b\):
\[ (x+a)(x+b) = c^2 \]
\[ x^2 + (a+b)x + ab - c^2 = 0 \]
This is now a standard quadratic in \(x\).

Step 4: Use the condition on the roots.
For a quadratic \(x^2 + Px + Q = 0\), the sum of the roots is \(-P\). Here the roots are equal in magnitude but opposite in sign, which means if one root is \(r\), the other is \(-r\). Their sum is then \(r + (-r) = 0\). So the sum of the roots of our quadratic must be zero:
\[ -(a+b) = 0 \]

Step 5: Solve for the required condition.
\[ a+b = 0 \]

Final Answer:
The roots are equal in magnitude and opposite in sign exactly when \(a+b=0\). \[ \boxed{a+b=0} \]