Q181 / 270 · NDA 2025 (Phase 2) · Mathematics
For the following two (02) items:
Consider the equation $abx^2 + bcx + ca = cax^2 + abx + bc$.
If the roots of the equation are equal, then which one of the following is correct?
Consider the equation $abx^2 + bcx + ca = cax^2 + abx + bc$.
If the roots of the equation are equal, then which one of the following is correct?
- A$ac = b^2$
- B$a + c = 2b$
- C$\dfrac{1}{a} + \dfrac{1}{c} = \dfrac{1}{2b}$
- Correct answerD$\dfrac{1}{a} + \dfrac{1}{c} = \dfrac{2}{b}$
Solution
The given equation is $abx^2 + bcx + ca = cax^2 + abx + bc$.
Rearranging the terms, we get:
$(ab - ca)x^2 + (bc - ab)x + (ca - bc) = 0$
$a(b - c)x^2 + b(c - a)x + c(a - b) = 0$
The sum of the coefficients is:
$a(b - c) + b(c - a) + c(a - b) = ab - ac + bc - ab + ac - bc = 0$
Since the sum of the coefficients is zero, one root of the equation is $x = 1$.
It is given that the roots of the equation are equal, so the other root must also be $x = 1$.
The product of the roots is given by $\dfrac{c(a - b)}{a(b - c)}$.
Equating the product of the roots to $1 \times 1 = 1$:
$\dfrac{c(a - b)}{a(b - c)} = 1$
$ca - cb = ab - ac$
$2ca = ab + bc$
Dividing both sides by $abc$, we get:
$\dfrac{2}{b} = \dfrac{1}{c} + \dfrac{1}{a}$
Answer: $\dfrac{1}{a} + \dfrac{1}{c} = \dfrac{2}{b}$
Rearranging the terms, we get:
$(ab - ca)x^2 + (bc - ab)x + (ca - bc) = 0$
$a(b - c)x^2 + b(c - a)x + c(a - b) = 0$
The sum of the coefficients is:
$a(b - c) + b(c - a) + c(a - b) = ab - ac + bc - ab + ac - bc = 0$
Since the sum of the coefficients is zero, one root of the equation is $x = 1$.
It is given that the roots of the equation are equal, so the other root must also be $x = 1$.
The product of the roots is given by $\dfrac{c(a - b)}{a(b - c)}$.
Equating the product of the roots to $1 \times 1 = 1$:
$\dfrac{c(a - b)}{a(b - c)} = 1$
$ca - cb = ab - ac$
$2ca = ab + bc$
Dividing both sides by $abc$, we get:
$\dfrac{2}{b} = \dfrac{1}{c} + \dfrac{1}{a}$
Answer: $\dfrac{1}{a} + \dfrac{1}{c} = \dfrac{2}{b}$
Hence, the correct answer is option: 'D'.