CAT 2023 Slot 1 QA Question 7

Type-in-the-answer (no negative marking) · Algebra · Polynomials · Try it, then check the answer and solution below.

The equation

x^3 + (2r + 1)x^2 + (4r - 1)x + 2 = 0

has - 2

as one of the roots. If the other two roots are real, then the minimum possible non-negative integer value of

r$ is

TITA Answer:

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Official Convenor Answer Key: 2

Answer and solution

Answer: 2

Given the cubic equation

x^3 + (2r + 1)x^2 + (4r - 1)x + 2 = 0

with - 2

as one of the roots, we need to find the minimum possible non-negative integer value of

r$ such that the other two roots are real.

Factor the cubic equation:

Since - 2

is a root,

(x + 2)

is a factor. Using synthetic division, we divide the cubic polynomial by

(x + 2)$:

\[ \begin{array}{r|rrrr} -2 & 1 & 2r + 1 & 4r - 1 & 2 \\ & & -2 & -4r + 2 & -2 \\ \hline & 1 & 2r - 1 & 0 & 0 \\ \end{array

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