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IBPS Clerk Mains 29 Nov 2025 Paper

Question Numbers: 130-131

An equation is given as:

\frac{2x^3 - 11x^2 + px}{x} = 0

One of the roots of the given equation is 3.

Question : 131 of 155

Marks: +1, -0

Find the difference between the roots of the given equation?

Solution:

Concept:

Use the given root to find the unknown coefficient

p

, then factor the quadratic and find the difference of its roots.

Explanation:

Given equation:

\frac{2x^3 - 11x^2 + px}{x} = 0

For

x \neq 0

, it simplifies to

2x^2 - 11x + p = 0

One root is

x = 3

. Substitute it:

2(3)^2 - 11(3) + p = 0

18 - 33 + p = 0

So,

p = 15

The equation becomes

2x^2 - 11x + 15 = 0

Factorize:

2x^2 - 5x - 6x + 15 = 0

(2x - 5)(x - 3) = 0

Thus, the roots are

x = 3

and

x = \frac{5}{2} = 2.5

Difference between the roots:

3 - 2.5 = 0.5 = \frac{1}{2}

Answer:

Option D:

\frac{1}{2}

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