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Telangana TET Paper 1 Exam Model Paper 4

Section: Mathematics

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Question : 111 of 150

Marks: +1, -0

The cubic equation whose roots are 1, 2, 3, is :

$x^{3} - 6x^{2} + 11x - 6 = 0$
$x3 + 6x2 - 11x - 6 = 0$
$x3 - 6x2 - 11x + 6 = 0$
$x3 + 6x2 + 11x - 6 = 0$

Solution:

Concept:

A cubic equation with given roots

\alpha, \beta, \gamma

is

x^3 - (\alpha+\beta+\gamma)x^2 + (\alpha\beta+\beta\gamma+\gamma\alpha)x - \alpha\beta\gamma = 0

.

Explanation:

Let the roots be 1, 2, and 3.

Sum of roots:

1+2+3 = 6

.

Sum of products taken two at a time:

(1\times2)+(2\times3)+(1\times3)=2+6+3=11

.

Product of all three:

1\times2\times3=6

.

Substitute into the formula:

x^3 - 6x^2 + 11x - 6 = 0

.

Answer:

Option A:

x^{3} - 6x^{2} + 11x - 6 = 0

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