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: Understanding the Concept.
Roots equal in magnitude but opposite in sign is a special condition on a quadratic equation: it always means the coefficient of $x$ in the quadratic must vanish, because the sum of such roots is zero, and the sum of roots equals minus the coefficient of $x$ once the quadratic is written with leading coefficient 1. So the real task is just to convert the given fractional equation into a plain quadratic in $x$ and read off when that middle coefficient is zero.

Step 2: Key Formula or Approach.
Start directly from
\[ \frac{x+a}{x+a+c} + \frac{x+b}{x+b+c} = 1 \]
Bring the left side to a common denominator $(x+a+c)(x+b+c)$:
\[ (x+a)(x+b+c) + (x+b)(x+a+c) = (x+a+c)(x+b+c) \]

Step 3: Detailed Explanation.
Expand the left side term by term:
$(x+a)(x+b+c) = x^2 + (a+b+c)x + ab+ac$
$(x+b)(x+a+c) = x^2 + (a+b+c)x + ab+bc$
Adding these: $2x^2 + 2(a+b+c)x + 2ab + ac + bc$.
Now expand the right side: $(x+a+c)(x+b+c) = x^2 + (a+b+2c)x + ab+ac+bc+c^2$.
Setting left equal to right and moving everything to one side:
\[ 2x^2 + 2(a+b+c)x + 2ab+ac+bc - x^2 - (a+b+2c)x - ab-ac-bc-c^2 = 0 \]
\[ x^2 + (a+b)x + ab - c^2 = 0 \]
The extra $c$ terms in the middle coefficient cancel to leave just $a+b$, and the constants simplify to $ab-c^2$. This matches the same reduced quadratic found through substitution, which is a good check that the algebra is consistent.

Step 4: Final Answer.
For the two roots of $x^2+(a+b)x+(ab-c^2)=0$ to be equal in size but opposite in sign, their sum must be zero. Since the sum of the roots is $-(a+b)$, we need
\[ a+b = 0 \]
This is the required condition. \[ \boxed{a+b=0} \]