Question:medium

Ujakar and Keshab attempted to solve a quadratic equation. Ujakar made a mistake in writing down the constant term. He ended up with the roots (4, 3). Keshab made a mistake in writing down the coefficient of x. He got the roots as (3, 2). What will be the exact roots of the orginal quadratic equation?

Updated On: Jan 13, 2026
  • (6,1)
  • (–3, –4)
  • (4,3)
  • (–4, –3)
  • (5,–3)
Show Solution

The Correct Option is A

Solution and Explanation

The correct answer is option (A):
(6,1)

Let the original quadratic equation be ax^2 + bx + c = 0.

Ujakar made a mistake in the constant term. This means the coefficient of x (which is b) and the coefficient of x^2 (which is a) were correct. He found roots 4 and 3. For a quadratic equation with roots r1 and r2, the equation can be written as k(x - r1)(x - r2) = 0, where k is a constant. Therefore, Ujakar's equation is of the form k(x - 4)(x - 3) = 0. Expanding this gives k(x^2 - 7x + 12) = 0. Since Ujakar's constant term was wrong, we can assume that the x^2 - 7x part is correct relative to the original. This means that if we take a=1 (we can assume this because we only care about the ratio of coefficients and the solutions), then the correct original equation would have the form x^2 - 7x + c = 0. Therefore, the sum of the roots is -(-7)/1 = 7.

Keshab made a mistake in the coefficient of x. This means the constant term (which is c) and the coefficient of x^2 (which is a) were correct. He got the roots as 3 and 2. Thus, Keshab's equation is k(x - 3)(x - 2) = 0. Expanding this gives k(x^2 - 5x + 6) = 0. Since Keshab's x coefficient was wrong, the constant term part would be correct. We can assume a=1. This would imply the correct original equation would have the form x^2 + bx + 6 = 0. The product of the roots is the constant term/leading coefficient, which is 6/1 = 6.

So we have an equation of the form x^2 - 7x + 6 = 0.
To find the roots, we factor this quadratic equation:
(x - 6)(x - 1) = 0
Therefore, the roots are x = 6 and x = 1.
The correct roots of the original quadratic equation are (6, 1).
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