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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 : 130 of 155

Marks: +1, -0

A new equation is given:

2x^{2} - 9x + 10 = 0

Solution:

Concept:

Find the value of

p

from the earlier equation, then compare it with the larger root of the given equation.

Explanation:

Given equation:

2x^2 - 9x + 10 = 0

Factorise:

2x^2 - 9x + 10 = (2x - 5)(x - 2)

So,

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

x = 2

Hence, the larger root is

2.5

From the earlier equation,

2x^2 - 11x + p = 0

has

x = 3

as a root.

Substitute

x = 3

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

18 - 33 + p = 0 \Rightarrow p = 15

Required difference =

|p - \text{larger root}| = |15 - 2.5| = 12.5

Answer:

12.5

(Option B).

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